Chapter 7 Integrals

We need to find a function whose derivative is the given function.

(i) $\cos 2x$

We know that the derivative of $\sin x$ is $\cos x$.

Consider the derivative of $\sin 2x$.

$\frac{d}{dx} (\sin 2x) = \cos 2x \cdot \frac{d}{dx}(2x) = \cos 2x \cdot 2 = 2 \cos 2x$

To get $\cos 2x$, we need to multiply $\sin 2x$ by $\frac{1}{2}$ before differentiating.

$\frac{d}{dx} \left(\frac{1}{2} \sin 2x\right) = \frac{1}{2} \frac{d}{dx} (\sin 2x) = \frac{1}{2} (2 \cos 2x) = \cos 2x$

Therefore, an anti derivative of $\cos 2x$ is $\mathbf{\frac{1}{2} \sin 2x}$.

(ii) $3x^2 + 4x^3$

We need to find a function whose derivative is $3x^2 + 4x^3$.

We know that the derivative of $x^n$ is $nx^{n-1}$. Conversely, the anti derivative of $x^m$ is $\frac{x^{m+1}}{m+1}$.

For the term $3x^2$, we look for a function whose derivative is $x^2$. This is $\frac{x^{2+1}}{2+1} = \frac{x^3}{3}$.

So, the anti derivative of $3x^2$ is $3 \cdot \frac{x^3}{3} = x^3$.

For the term $4x^3$, we look for a function whose derivative is $x^3$. This is $\frac{x^{3+1}}{3+1} = \frac{x^4}{4}$.

So, the anti derivative of $4x^3$ is $4 \cdot \frac{x^4}{4} = x^4$.

By the linearity of differentiation, the anti derivative of the sum is the sum of the anti derivatives.

Thus, the anti derivative of $3x^2 + 4x^3$ is the sum of the anti derivatives of $3x^2$ and $4x^3$.

An anti derivative is $x^3 + x^4$.

Let's verify by differentiation:

$\frac{d}{dx}(x^3 + x^4) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^4) = 3x^2 + 4x^3$.

Therefore, an anti derivative of $3x^2 + 4x^3$ is $\mathbf{x^3 + x^4}$.

(iii) $\frac{1}{x}$, $x \neq 0$

We need to find a function whose derivative is $\frac{1}{x}$.

We know that the derivative of $\ln |x|$ is $\frac{1}{x}$ for $x \neq 0$.

$\frac{d}{dx} (\ln |x|) = \frac{1}{x}$, for $x \neq 0$.

Therefore, an anti derivative of $\frac{1}{x}$ is $\mathbf{\ln |x|}$.

We use the properties of integration and the standard integration formulas.

(i) $\int \frac{x^3 − 1}{x^2} \;dx$

First, we can rewrite the integrand by dividing each term in the numerator by the denominator:

$\frac{x^3 - 1}{x^2} = \frac{x^3}{x^2} - \frac{1}{x^2} = x - x^{-2}$

Now, we integrate term by term:

$\int (x - x^{-2}) \;dx = \int x \;dx - \int x^{-2} \;dx$

Using the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$\int x \;dx = \frac{x^{1+1}}{1+1} + C_1 = \frac{x^2}{2} + C_1$

$\int x^{-2} \;dx = \frac{x^{-2+1}}{-2+1} + C_2 = \frac{x^{-1}}{-1} + C_2 = -\frac{1}{x} + C_2$

Combining these, we get:

$\int \frac{x^3 − 1}{x^2} \;dx = \frac{x^2}{2} - \left(-\frac{1}{x}\right) + C = \frac{x^2}{2} + \frac{1}{x} + C$

where $C = C_1 - C_2$ is the constant of integration.

Therefore, the integral is $\mathbf{\frac{x^2}{2} + \frac{1}{x} + C}$.

(ii) $\int (x^{\frac{2}{3}} + 1) \;dx$

We can integrate term by term:

$\int (x^{\frac{2}{3}} + 1) \;dx = \int x^{\frac{2}{3}} \;dx + \int 1 \;dx$

Using the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$:

$\int x^{\frac{2}{3}} \;dx = \frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1} + C_1 = \frac{x^{\frac{5}{3}}}{\frac{5}{3}} + C_1 = \frac{3}{5} x^{\frac{5}{3}} + C_1$

$\int 1 \;dx = \int x^0 \;dx = \frac{x^{0+1}}{0+1} + C_2 = x + C_2$

Combining these, we get:

$\int (x^{\frac{2}{3}} + 1) \;dx = \frac{3}{5} x^{\frac{5}{3}} + x + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral is $\mathbf{\frac{3}{5} x^{\frac{5}{3}} + x + C}$.

(iii) $\int (x^{\frac{3}{2}} + 2e^x − \frac{1}{x}) \;dx$

We can integrate term by term:

$\int (x^{\frac{3}{2}} + 2e^x − \frac{1}{x}) \;dx = \int x^{\frac{3}{2}} \;dx + \int 2e^x \;dx − \int \frac{1}{x} \;dx$

Using the standard integration formulas:

$\int x^{\frac{3}{2}} \;dx = \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} + C_1 = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + C_1 = \frac{2}{5} x^{\frac{5}{2}} + C_1$

$\int 2e^x \;dx = 2 \int e^x \;dx = 2e^x + C_2$

$\int \frac{1}{x} \;dx = \ln |x| + C_3$ (for $x \neq 0$)

Combining these, we get:

$\int (x^{\frac{3}{2}} + 2e^x − \frac{1}{x}) \;dx = \frac{2}{5} x^{\frac{5}{2}} + 2e^x - \ln |x| + C$

where $C = C_1 + C_2 - C_3$ is the constant of integration.

Therefore, the integral is $\mathbf{\frac{2}{5} x^{\frac{5}{2}} + 2e^x - \ln |x| + C}$.

We use the linearity property of integrals and standard trigonometric integral formulas.

(i) $\int (\sin x + \cos x ) \;dx$

Using the linearity property, we can write:

$\int (\sin x + \cos x) \;dx = \int \sin x \;dx + \int \cos x \;dx$

We know that $\int \sin x \;dx = -\cos x + C_1$ and $\int \cos x \;dx = \sin x + C_2$.

So, $\int (\sin x + \cos x) \;dx = -\cos x + \sin x + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral is $\mathbf{\sin x - \cos x + C}$.

(ii) $\int \text{cosec}\; x (\text{cosec}\; x + \cot x) \;dx$

First, we expand the integrand:

$\text{cosec}\; x (\text{cosec}\; x + \cot x) = \text{cosec}^2 x + \text{cosec}\; x \cot x$

Now, we integrate term by term:

$\int (\text{cosec}^2 x + \text{cosec}\; x \cot x) \;dx = \int \text{cosec}^2 x \;dx + \int \text{cosec}\; x \cot x \;dx$

We know that $\int \text{cosec}^2 x \;dx = -\cot x + C_1$ and $\int \text{cosec}\; x \cot x \;dx = -\text{cosec}\; x + C_2$.

So, $\int \text{cosec}\; x (\text{cosec}\; x + \cot x) \;dx = -\cot x - \text{cosec}\; x + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral is $\mathbf{-\cot x - \text{cosec}\; x + C}$.

(iii) $\int \frac{1 − \sin x}{\cos^2 x} \;dx$

We can rewrite the integrand by splitting the fraction:

$\frac{1 - \sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}$

Using trigonometric identities, we have $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x$.

So the integral becomes:

$\int (\sec^2 x - \sec x \tan x) \;dx$

Using the linearity property, we integrate term by term:

$\int (\sec^2 x - \sec x \tan x) \;dx = \int \sec^2 x \;dx - \int \sec x \tan x \;dx$

We know that $\int \sec^2 x \;dx = \tan x + C_1$ and $\int \sec x \tan x \;dx = \sec x + C_2$.

So, $\int \frac{1 − \sin x}{\cos^2 x} \;dx = \tan x - \sec x + C$

where $C = C_1 - C_2$ is the constant of integration.

Therefore, the integral is $\mathbf{\tan x - \sec x + C}$.

We are given the function $f(x) = 4x^3 - 6$ and we need to find its anti derivative $F(x)$ such that $F(0) = 3$.

The anti derivative $F(x)$ is the indefinite integral of $f(x)$.

$F(x) = \int f(x) \;dx = \int (4x^3 - 6) \;dx$

Using the linearity of the integral and the power rule ($\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ and $\int k \;dx = kx + C$):

$F(x) = \int 4x^3 \;dx - \int 6 \;dx$

$F(x) = 4 \int x^3 \;dx - 6 \int 1 \;dx$

$F(x) = 4 \left(\frac{x^{3+1}}{3+1}\right) - 6(x) + C$

$F(x) = 4 \left(\frac{x^4}{4}\right) - 6x + C$

$F(x) = x^4 - 6x + C$

We are given the condition $F(0) = 3$. We can use this to find the value of the constant of integration, $C$.

Substitute $x=0$ into the expression for $F(x)$:

$F(0) = (0)^4 - 6(0) + C$

$3 = 0 - 0 + C$

$3 = C$

Now, substitute the value of $C$ back into the expression for $F(x)$.

$F(x) = x^4 - 6x + 3$

Therefore, the anti derivative F of f satisfying the given condition is $\mathbf{F(x) = x^4 - 6x + 3}$.

We need to find a function whose derivative with respect to $x$ is $\sin 2x$.

We know that the derivative of $\cos x$ is $-\sin x$.

Let's consider the derivative of $\cos 2x$.

$\frac{d}{dx}(\cos 2x) = -\sin 2x \cdot \frac{d}{dx}(2x) = -\sin 2x \cdot 2 = -2 \sin 2x$

We want the derivative to be $\sin 2x$, which is $-\frac{1}{2}$ times $-2 \sin 2x$.

So, let's consider the derivative of $-\frac{1}{2} \cos 2x$.

$\frac{d}{dx}\left(-\frac{1}{2} \cos 2x\right) = -\frac{1}{2} \frac{d}{dx}(\cos 2x) = -\frac{1}{2} (-2 \sin 2x) = \sin 2x$

Thus, an anti derivative of $\sin 2x$ is $-\frac{1}{2} \cos 2x$.

Therefore, an anti derivative of $\sin 2x$ is $\mathbf{-\frac{1}{2} \cos 2x}$.

We need to find a function whose derivative with respect to $x$ is $\cos 3x$.

We know that the derivative of $\sin x$ is $\cos x$.

Let's consider the derivative of $\sin 3x$.

$\frac{d}{dx}(\sin 3x) = \cos 3x \cdot \frac{d}{dx}(3x) = \cos 3x \cdot 3 = 3 \cos 3x$

We want the derivative to be $\cos 3x$, which is $\frac{1}{3}$ times $3 \cos 3x$.

So, let's consider the derivative of $\frac{1}{3} \sin 3x$.

$\frac{d}{dx}\left(\frac{1}{3} \sin 3x\right) = \frac{1}{3} \frac{d}{dx}(\sin 3x) = \frac{1}{3} (3 \cos 3x) = \cos 3x$

Thus, an anti derivative of $\cos 3x$ is $\frac{1}{3} \sin 3x$.

Therefore, an anti derivative of $\cos 3x$ is $\mathbf{\frac{1}{3} \sin 3x}$.

We need to find a function whose derivative with respect to $x$ is $e^{2x}$.

We know that the derivative of $e^x$ is $e^x$.

Let's consider the derivative of $e^{2x}$.

$\frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = e^{2x} \cdot 2 = 2e^{2x}$

We want the derivative to be $e^{2x}$, which is $\frac{1}{2}$ times $2e^{2x}$.

So, let's consider the derivative of $\frac{1}{2} e^{2x}$.

$\frac{d}{dx}\left(\frac{1}{2} e^{2x}\right) = \frac{1}{2} \frac{d}{dx}(e^{2x}) = \frac{1}{2} (2e^{2x}) = e^{2x}$

Thus, an anti derivative of $e^{2x}$ is $\frac{1}{2} e^{2x}$.

Therefore, an anti derivative of $e^{2x}$ is $\mathbf{\frac{1}{2} e^{2x}}$.

We need to find a function whose derivative with respect to $x$ is $(ax+b)^2$.

We know that $\frac{d}{dx}(x^n) = nx^{n-1}$. For a linear function inside the power, we use the chain rule: $\frac{d}{dx}((ax+b)^n) = n(ax+b)^{n-1} \cdot a$.

Let's consider the derivative of $(ax+b)^3$.

$\frac{d}{dx}((ax+b)^3) = 3(ax+b)^{3-1} \cdot \frac{d}{dx}(ax+b)$

$\frac{d}{dx}((ax+b)^3) = 3(ax+b)^2 \cdot a$

$\frac{d}{dx}((ax+b)^3) = 3a(ax+b)^2$

We want the derivative to be $(ax+b)^2$. Currently, we have $3a(ax+b)^2$.

To get $(ax+b)^2$, we need to multiply the function $(ax+b)^3$ by $\frac{1}{3a}$ before differentiating.

Let's consider the derivative of $\frac{1}{3a}(ax+b)^3$.

$\frac{d}{dx}\left(\frac{1}{3a}(ax+b)^3\right) = \frac{1}{3a} \frac{d}{dx}((ax+b)^3)$

$\frac{d}{dx}\left(\frac{1}{3a}(ax+b)^3\right) = \frac{1}{3a} (3a(ax+b)^2)$

$\frac{d}{dx}\left(\frac{1}{3a}(ax+b)^3\right) = (ax+b)^2$

Thus, an anti derivative of $(ax+b)^2$ is $\frac{1}{3a}(ax+b)^3$.

Therefore, an anti derivative of $(ax+b)^2$ is $\mathbf{\frac{1}{3a}(ax+b)^3}$.

We need to find a function whose derivative with respect to $x$ is $\sin^2 x - 4e^{3x}$. This can be done by finding anti-derivatives for each term separately.

First, consider the term $4e^{3x}$. We know that $\frac{d}{dx}(e^{kx}) = ke^{kx}$.

Let's guess a function of the form $Ae^{3x}$.

$\frac{d}{dx}(Ae^{3x}) = A \cdot \frac{d}{dx}(e^{3x}) = A \cdot 3e^{3x} = 3Ae^{3x}$.

We want this to be equal to $4e^{3x}$. So, $3A = 4$, which means $A = \frac{4}{3}$.

Thus, an anti-derivative of $4e^{3x}$ is $\frac{4}{3}e^{3x}$.

Next, consider the term $\sin^2 x$. We use the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$, which implies $\sin^2 x = \frac{1 - \cos 2x}{2} = \frac{1}{2} - \frac{1}{2}\cos 2x$.

We need to find an anti-derivative of $\frac{1}{2} - \frac{1}{2}\cos 2x$. This can be done by finding anti-derivatives for $\frac{1}{2}$ and $-\frac{1}{2}\cos 2x$ separately.

For the term $\frac{1}{2}$, we guess a function of the form $Bx$.

$\frac{d}{dx}(Bx) = B$. We want this to be $\frac{1}{2}$, so $B = \frac{1}{2}$.

An anti-derivative of $\frac{1}{2}$ is $\frac{1}{2}x$.

For the term $-\frac{1}{2}\cos 2x$, we know $\frac{d}{dx}(\sin kx) = k\cos kx$. Let's guess a function of the form $C\sin 2x$.

$\frac{d}{dx}(C\sin 2x) = C \cdot \frac{d}{dx}(\sin 2x) = C \cdot (2\cos 2x) = 2C\cos 2x$.

We want this to be $-\frac{1}{2}\cos 2x$. So, $2C = -\frac{1}{2}$, which means $C = -\frac{1}{4}$.

An anti-derivative of $-\frac{1}{2}\cos 2x$ is $-\frac{1}{4}\sin 2x$.

Combining these, an anti-derivative of $\sin^2 x$ is the sum of the anti-derivatives of its components: $\frac{1}{2}x - \frac{1}{4}\sin 2x$.

Finally, the anti-derivative of $\sin^2 x - 4e^{3x}$ is the anti-derivative of $\sin^2 x$ minus the anti-derivative of $4e^{3x}$.

Anti-derivative $= \left(\frac{1}{2}x - \frac{1}{4}\sin 2x\right) - \left(\frac{4}{3}e^{3x}\right)$.

Therefore, an anti derivative of $\sin^2 x – 4e^{3x}$ is $\mathbf{\frac{1}{2}x - \frac{1}{4}\sin 2x - \frac{4}{3}e^{3x}}$.

We need to find the integral of the function $4e^{3x} + 1$ with respect to $x$.

Using the linearity property of integration, we can separate the integral into two parts:

$\int (4e^{3x} + 1)\;dx = \int 4e^{3x} \;dx + \int 1 \;dx$

For the first integral, $\int 4e^{3x} \;dx$, we can take the constant 4 outside the integral:

$4 \int e^{3x} \;dx$

Using the standard integral formula $\int e^{ax} \;dx = \frac{1}{a} e^{ax} + C$, with $a=3$, we get:

$4 \left(\frac{1}{3} e^{3x}\right) + C_1 = \frac{4}{3} e^{3x} + C_1$

For the second integral, $\int 1 \;dx$, which is the integral of a constant, we get:

$\int 1 \;dx = x + C_2$

Combining the results from both parts, we get the final integral:

$\int (4e^{3x} + 1)\;dx = \left(\frac{4}{3} e^{3x} + C_1\right) + (x + C_2)$

$\int (4e^{3x} + 1)\;dx = \frac{4}{3} e^{3x} + x + C_1 + C_2$

Let $C = C_1 + C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{4}{3} e^{3x} + x + C}$.

We need to find the integral of the function $x^2 \left(1 - \frac{1}{x^2}\right)$ with respect to $x$.

First, simplify the integrand by multiplying $x^2$ through the parenthesis:

$x^2 \left(1 - \frac{1}{x^2}\right) = x^2 \cdot 1 - x^2 \cdot \frac{1}{x^2} = x^2 - 1$

Now, we need to find the integral of the simplified expression:

$\int (x^2 - 1) \;dx$

Using the linearity property of integration, we can integrate term by term:

$\int (x^2 - 1) \;dx = \int x^2 \;dx - \int 1 \;dx$

Using the power rule for integration $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), and $\int k \;dx = kx + C$ for a constant $k$:

$\int x^2 \;dx = \frac{x^{2+1}}{2+1} + C_1 = \frac{x^3}{3} + C_1$

$\int 1 \;dx = x + C_2$

Combining these results, we get:

$\int (x^2 - 1) \;dx = \left(\frac{x^3}{3} + C_1\right) - (x + C_2) = \frac{x^3}{3} - x + (C_1 - C_2)$

Let $C = C_1 - C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{x^3}{3} - x + C}$.

We need to find the integral of the polynomial function $(ax^2 + bx + c)$ with respect to $x$, where $a$, $b$, and $c$ are constants.

Using the linearity property of integration, we can integrate each term separately:

$\int (ax^2 + bx + c) \;dx = \int ax^2 \;dx + \int bx \;dx + \int c \;dx$

For the first term, $\int ax^2 \;dx$, we take the constant $a$ outside and use the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$:

$\int ax^2 \;dx = a \int x^2 \;dx = a \left(\frac{x^{2+1}}{2+1}\right) + C_1 = a \frac{x^3}{3} + C_1$

For the second term, $\int bx \;dx$, we take the constant $b$ outside and use the power rule:

$\int bx \;dx = b \int x^1 \;dx = b \left(\frac{x^{1+1}}{1+1}\right) + C_2 = b \frac{x^2}{2} + C_2$

For the third term, $\int c \;dx$, which is the integral of a constant, we use the formula $\int k \;dx = kx + C$:

$\int c \;dx = cx + C_3$

Combining the results from all terms, we get the final integral:

$\int (ax^2 + bx + c) \;dx = \left(a \frac{x^3}{3} + C_1\right) + \left(b \frac{x^2}{2} + C_2\right) + (cx + C_3)$

$\int (ax^2 + bx + c) \;dx = a \frac{x^3}{3} + b \frac{x^2}{2} + cx + (C_1 + C_2 + C_3)$

Let $C = C_1 + C_2 + C_3$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + C}$.

We need to find the integral of the function $2x^2 + e^x$ with respect to $x$.

Using the linearity property of integration, we can integrate each term separately:

$\int (2x^2 + e^x) \;dx = \int 2x^2 \;dx + \int e^x \;dx$

For the first term, $\int 2x^2 \;dx$, we take the constant 2 outside and use the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$:

$\int 2x^2 \;dx = 2 \int x^2 \;dx = 2 \left(\frac{x^{2+1}}{2+1}\right) + C_1 = 2 \frac{x^3}{3} + C_1$

For the second term, $\int e^x \;dx$, we use the standard integral formula $\int e^x \;dx = e^x + C_2$:

$\int e^x \;dx = e^x + C_2$

Combining the results from both terms, we get the final integral:

$\int (2x^2 + e^x) \;dx = \left(\frac{2x^3}{3} + C_1\right) + (e^x + C_2) = \frac{2x^3}{3} + e^x + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{2}{3}x^3 + e^x + C}$.

We need to find the integral of the function $\left( \sqrt{x} − \frac{1}{\sqrt{x}} \right)^2$ with respect to $x$.

First, simplify the integrand by expanding the square:

$\left( \sqrt{x} − \frac{1}{\sqrt{x}} \right)^2 = (\sqrt{x})^2 - 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2$

$\left( \sqrt{x} − \frac{1}{\sqrt{x}} \right)^2 = x - 2(1) + \frac{1}{x}$

$\left( \sqrt{x} − \frac{1}{\sqrt{x}} \right)^2 = x - 2 + \frac{1}{x}$

Now, we need to find the integral of the simplified expression:

$\int \left( x - 2 + \frac{1}{x} \right) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int x \;dx - \int 2 \;dx + \int \frac{1}{x} \;dx$

Using the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$) and the integral of $\frac{1}{x}$:

$\int x \;dx = \frac{x^{1+1}}{1+1} + C_1 = \frac{x^2}{2} + C_1$

$\int 2 \;dx = 2x + C_2$

$\int \frac{1}{x} \;dx = \ln |x| + C_3$ (for $x \neq 0$)

Combining the results from all terms, we get the final integral:

$\int \left( x - 2 + \frac{1}{x} \right) \;dx = \left(\frac{x^2}{2} + C_1\right) - (2x + C_2) + (\ln |x| + C_3)$

$\int \left( x - 2 + \frac{1}{x} \right) \;dx = \frac{x^2}{2} - 2x + \ln |x| + (C_1 - C_2 + C_3)$

Let $C = C_1 - C_2 + C_3$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{x^2}{2} - 2x + \ln |x| + C}$.

We need to find the integral of the function $\frac{x^3 + 5x^2 − 4}{x^2}$ with respect to $x$.

First, simplify the integrand by dividing each term in the numerator by $x^2$:

$\frac{x^3 + 5x^2 - 4}{x^2} = \frac{x^3}{x^2} + \frac{5x^2}{x^2} - \frac{4}{x^2}$

$\frac{x^3 + 5x^2 - 4}{x^2} = x + 5 - 4x^{-2}$

Now, we need to find the integral of the simplified expression:

$\int (x + 5 - 4x^{-2}) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int x \;dx + \int 5 \;dx - \int 4x^{-2} \;dx$

Using the power rule for integration $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$) and $\int k \;dx = kx + C$ for a constant $k$:

$\int x \;dx = \frac{x^{1+1}}{1+1} + C_1 = \frac{x^2}{2} + C_1$

$\int 5 \;dx = 5x + C_2$

$\int 4x^{-2} \;dx = 4 \int x^{-2} \;dx = 4 \left(\frac{x^{-2+1}}{-2+1}\right) + C_3 = 4 \left(\frac{x^{-1}}{-1}\right) + C_3 = -4x^{-1} + C_3 = -\frac{4}{x} + C_3$

Combining these results, we get the final integral:

$\int \left( x + 5 - 4x^{-2} \right) \;dx = \left(\frac{x^2}{2} + C_1\right) + (5x + C_2) - \left(-\frac{4}{x} + C_3\right)$

$\int \left( x + 5 - 4x^{-2} \right) \;dx = \frac{x^2}{2} + 5x + \frac{4}{x} + (C_1 + C_2 - C_3)$

Let $C = C_1 + C_2 - C_3$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{x^2}{2} + 5x + \frac{4}{x} + C}$.

We need to find the integral of the function $\frac{x^3 + 3x + 4}{\sqrt{x}}$ with respect to $x$.

First, simplify the integrand by dividing each term in the numerator by $\sqrt{x} = x^{1/2}$:

$\frac{x^3 + 3x + 4}{\sqrt{x}} = \frac{x^3}{x^{1/2}} + \frac{3x}{x^{1/2}} + \frac{4}{x^{1/2}}$

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{x^3}{x^{1/2}} = x^{3 - \frac{1}{2}} = x^{\frac{6-1}{2}} = x^{\frac{5}{2}}$

$\frac{3x}{x^{1/2}} = 3x^{1 - \frac{1}{2}} = 3x^{\frac{2-1}{2}} = 3x^{\frac{1}{2}}$

$\frac{4}{x^{1/2}} = 4x^{-\frac{1}{2}}$

So, the simplified integrand is $x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}$.

Now, we need to find the integral of the simplified expression:

$\int \left( x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int x^{\frac{5}{2}} \;dx + \int 3x^{\frac{1}{2}} \;dx + \int 4x^{-\frac{1}{2}} \;dx$

Using the power rule for integration $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$\int x^{\frac{5}{2}} \;dx = \frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1} + C_1 = \frac{x^{\frac{7}{2}}}{\frac{7}{2}} + C_1 = \frac{2}{7}x^{\frac{7}{2}} + C_1$

$\int 3x^{\frac{1}{2}} \;dx = 3 \int x^{\frac{1}{2}} \;dx = 3 \left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right) + C_2 = 3 \left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right) + C_2 = 3 \cdot \frac{2}{3} x^{\frac{3}{2}} + C_2 = 2x^{\frac{3}{2}} + C_2$

$\int 4x^{-\frac{1}{2}} \;dx = 4 \int x^{-\frac{1}{2}} \;dx = 4 \left(\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right) + C_3 = 4 \left(\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right) + C_3 = 4 \cdot 2 x^{\frac{1}{2}} + C_3 = 8x^{\frac{1}{2}} + C_3$

Combining these results, we get the final integral:

$\int \left( x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) \;dx = \left(\frac{2}{7}x^{\frac{7}{2}} + C_1\right) + \left(2x^{\frac{3}{2}} + C_2\right) + \left(8x^{\frac{1}{2}} + C_3\right)$

$\int \left( x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) \;dx = \frac{2}{7}x^{\frac{7}{2}} + 2x^{\frac{3}{2}} + 8x^{\frac{1}{2}} + (C_1 + C_2 + C_3)$

Let $C = C_1 + C_2 + C_3$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{2}{7}x^{\frac{7}{2}} + 2x^{\frac{3}{2}} + 8\sqrt{x} + C}$.

We need to find the integral of the function $\frac{x^3 − x^2 + x − 1}{x − 1}$ with respect to $x$.

First, simplify the integrand by factoring the numerator. Notice that we can group terms:

$x^3 - x^2 + x - 1 = x^2(x-1) + 1(x-1) = (x^2 + 1)(x-1)$

So, the integrand becomes:

$\frac{(x^2 + 1)(x-1)}{x-1}$

For $x \neq 1$, we can cancel the $(x-1)$ term:

$\frac{(x^2 + 1)\cancel{(x-1)}}{\cancel{x-1}} = x^2 + 1$

Now, we need to find the integral of the simplified expression:

$\int (x^2 + 1) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int x^2 \;dx + \int 1 \;dx$

Using the power rule for integration $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$) and $\int k \;dx = kx + C$ for a constant $k$:

$\int x^2 \;dx = \frac{x^{2+1}}{2+1} + C_1 = \frac{x^3}{3} + C_1$

$\int 1 \;dx = x + C_2$

Combining these results, we get the final integral:

$\int (x^2 + 1) \;dx = \left(\frac{x^3}{3} + C_1\right) + (x + C_2) = \frac{x^3}{3} + x + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{x^3}{3} + x + C}$.

We need to find the integral of the function $(1 - x) \sqrt{x}$ with respect to $x$.

First, simplify the integrand by multiplying $\sqrt{x}$ through the parenthesis:

$(1 - x) \sqrt{x} = 1 \cdot \sqrt{x} - x \cdot \sqrt{x}$

Recall that $\sqrt{x} = x^{1/2}$. So, $x \cdot \sqrt{x} = x^1 \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}$.

The simplified integrand is $x^{1/2} - x^{3/2}$.

Now, we need to find the integral of the simplified expression:

$\int (x^{1/2} - x^{3/2}) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int x^{1/2} \;dx - \int x^{3/2} \;dx$

Using the power rule for integration $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$\int x^{1/2} \;dx = \frac{x^{1/2+1}}{\frac{1}{2}+1} + C_1 = \frac{x^{3/2}}{\frac{3}{2}} + C_1 = \frac{2}{3}x^{3/2} + C_1$

$\int x^{3/2} \;dx = \frac{x^{3/2+1}}{\frac{3}{2}+1} + C_2 = \frac{x^{5/2}}{\frac{5}{2}} + C_2 = \frac{2}{5}x^{5/2} + C_2$

Combining these results, we get the final integral:

$\int (x^{1/2} - x^{3/2}) \;dx = \left(\frac{2}{3}x^{3/2} + C_1\right) - \left(\frac{2}{5}x^{5/2} + C_2\right)$

$\int (x^{1/2} - x^{3/2}) \;dx = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + (C_1 - C_2)$

Let $C = C_1 - C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{2}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} + C}$.

We need to find the integral of the function $\sqrt{x}(3x^2 + 2x + 3)$ with respect to $x$.

First, simplify the integrand by multiplying $\sqrt{x} = x^{1/2}$ by each term inside the parenthesis:

$\sqrt{x}(3x^2 + 2x + 3) = x^{1/2}(3x^2) + x^{1/2}(2x) + x^{1/2}(3)$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$:

$x^{1/2} \cdot 3x^2 = 3x^{1/2 + 2} = 3x^{1/2 + 4/2} = 3x^{5/2}$

$x^{1/2} \cdot 2x = 2x^{1/2 + 1} = 2x^{1/2 + 2/2} = 2x^{3/2}$

$x^{1/2} \cdot 3 = 3x^{1/2}$

The simplified integrand is $3x^{5/2} + 2x^{3/2} + 3x^{1/2}$.

Now, we need to find the integral of the simplified expression:

$\int \left( 3x^{5/2} + 2x^{3/2} + 3x^{1/2} \right) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int 3x^{5/2} \;dx + \int 2x^{3/2} \;dx + \int 3x^{1/2} \;dx$

Take the constants outside the integrals:

$3 \int x^{5/2} \;dx + 2 \int x^{3/2} \;dx + 3 \int x^{1/2} \;dx$

Using the power rule for integration $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$\int x^{5/2} \;dx = \frac{x^{5/2+1}}{5/2+1} + C_1 = \frac{x^{7/2}}{7/2} + C_1 = \frac{2}{7}x^{7/2} + C_1$

$\int x^{3/2} \;dx = \frac{x^{3/2+1}}{3/2+1} + C_2 = \frac{x^{5/2}}{5/2} + C_2 = \frac{2}{5}x^{5/2} + C_2$

$\int x^{1/2} \;dx = \frac{x^{1/2+1}}{1/2+1} + C_3 = \frac{x^{3/2}}{3/2} + C_3 = \frac{2}{3}x^{3/2} + C_3$

Substitute these results back into the integral expression:

$3 \left(\frac{2}{7}x^{7/2}\right) + 2 \left(\frac{2}{5}x^{5/2}\right) + 3 \left(\frac{2}{3}x^{3/2}\right) + C$

where $C = 3C_1 + 2C_2 + 3C_3$ is the constant of integration (or rather $C$ absorbs all individual constants).

Simplify the coefficients:

$\frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + \frac{6}{3}x^{3/2} + C$

$\frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C$

Therefore, the integral is $\mathbf{\frac{6}{7}x^{\frac{7}{2}} + \frac{4}{5}x^{\frac{5}{2}} + 2x^{\frac{3}{2}} + C}$.

We need to find the integral of the function $2x - 3 \cos x + e^x$ with respect to $x$.

Using the linearity property of integration, we integrate each term separately:

$\int (2x − 3 \cos x + e^x)\; dx = \int 2x \;dx - \int 3 \cos x \;dx + \int e^x \;dx$

For the first term, $\int 2x \;dx$, we take the constant 2 outside and use the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$:

$\int 2x \;dx = 2 \int x^1 \;dx = 2 \left(\frac{x^{1+1}}{1+1}\right) + C_1 = 2 \frac{x^2}{2} + C_1 = x^2 + C_1$

For the second term, $\int 3 \cos x \;dx$, we take the constant 3 outside and use the standard integral formula $\int \cos x \;dx = \sin x + C$:

$\int 3 \cos x \;dx = 3 \int \cos x \;dx = 3 (\sin x) + C_2 = 3 \sin x + C_2$

For the third term, $\int e^x \;dx$, we use the standard integral formula $\int e^x \;dx = e^x + C_3$:

$\int e^x \;dx = e^x + C_3$

Combining these results, we get the final integral:

$\int (2x − 3 \cos x + e^x)\; dx = (x^2 + C_1) - (3 \sin x + C_2) + (e^x + C_3)$

$\int (2x − 3 \cos x + e^x)\; dx = x^2 - 3 \sin x + e^x + (C_1 - C_2 + C_3)$

Let $C = C_1 - C_2 + C_3$ be the constant of integration.

Therefore, the integral is $\mathbf{x^2 - 3 \sin x + e^x + C}$.

We need to find the integral of the function $2x^2 - 3 \sin x + 5 \sqrt{x}$ with respect to $x$.

Using the linearity property of integration, we integrate each term separately:

$\int (2x^2 − 3 \sin x + 5 \sqrt{x}) \;dx = \int 2x^2 \;dx - \int 3 \sin x \;dx + \int 5 \sqrt{x} \;dx$

For the first term, $\int 2x^2 \;dx$, we take the constant 2 outside and use the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$:

$\int 2x^2 \;dx = 2 \int x^2 \;dx = 2 \left(\frac{x^{2+1}}{2+1}\right) + C_1 = 2 \frac{x^3}{3} + C_1 = \frac{2}{3}x^3 + C_1$

For the second term, $\int 3 \sin x \;dx$, we take the constant 3 outside and use the standard integral formula $\int \sin x \;dx = -\cos x + C$:

$\int 3 \sin x \;dx = 3 \int \sin x \;dx = 3 (-\cos x) + C_2 = -3 \cos x + C_2$

For the third term, $\int 5 \sqrt{x} \;dx$, we write $\sqrt{x}$ as $x^{1/2}$, take the constant 5 outside, and use the power rule:

$\int 5 \sqrt{x} \;dx = \int 5x^{1/2} \;dx = 5 \int x^{1/2} \;dx = 5 \left(\frac{x^{1/2+1}}{1/2+1}\right) + C_3 = 5 \left(\frac{x^{3/2}}{3/2}\right) + C_3 = 5 \cdot \frac{2}{3} x^{3/2} + C_3 = \frac{10}{3} x^{3/2} + C_3$

Combining these results, we get the final integral:

$\int (2x^2 − 3 \sin x + 5 \sqrt{x}) \;dx = \left(\frac{2}{3}x^3 + C_1\right) - (-3 \cos x + C_2) + \left(\frac{10}{3} x^{3/2} + C_3\right)$

$\int (2x^2 − 3 \sin x + 5 \sqrt{x}) \;dx = \frac{2}{3}x^3 + 3 \cos x + \frac{10}{3} x^{3/2} + (C_1 - C_2 + C_3)$

Let $C = C_1 - C_2 + C_3$ be the constant of integration.

Therefore, the integral is $\mathbf{\frac{2}{3}x^3 + 3 \cos x + \frac{10}{3} x^{\frac{3}{2}} + C}$.

We need to find the integral of the function $\sec x (\sec x + \tan x)$ with respect to $x$.

First, simplify the integrand by multiplying $\sec x$ through the parenthesis:

$\sec x (\sec x + \tan x) = \sec x \cdot \sec x + \sec x \cdot \tan x$

$\sec x (\sec x + \tan x) = \sec^2 x + \sec x \tan x$

Now, we need to find the integral of the simplified expression:

$\int (\sec^2 x + \sec x \tan x) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int \sec^2 x \;dx + \int \sec x \tan x \;dx$

Using the standard integral formulas:

$\int \sec^2 x \;dx = \tan x + C_1$

$\int \sec x \tan x \;dx = \sec x + C_2$

Combining these results, we get the final integral:

$\int (\sec^2 x + \sec x \tan x) \;dx = (\tan x + C_1) + (\sec x + C_2)$

$\int (\sec^2 x + \sec x \tan x) \;dx = \tan x + \sec x + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{\tan x + \sec x + C}$.

We need to find the integral of the function $\frac{\sec^2 x}{\text{cosec}^2 x}$ with respect to $x$.

First, simplify the integrand using trigonometric identities. Recall that $\sec x = \frac{1}{\cos x}$ and $\text{cosec}\; x = \frac{1}{\sin x}$.

$\sec^2 x = \frac{1}{\cos^2 x}$

$\text{cosec}^2 \;x = \frac{1}{\sin^2 x}$

So, the integrand is:

$\frac{\sec^2 x}{\text{cosec}^2 x} = \frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}} = \frac{1}{\cos^2 x} \cdot \frac{\sin^2 x}{1} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$

Now, we need to find the integral of $\tan^2 x$. We use the trigonometric identity $\tan^2 x = \sec^2 x - 1$.

$\int \tan^2 x \;dx = \int (\sec^2 x - 1) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int \sec^2 x \;dx - \int 1 \;dx$

Using the standard integral formulas:

$\int \sec^2 x \;dx = \tan x + C_1$

$\int 1 \;dx = x + C_2$

Combining these results, we get the final integral:

$\int \tan^2 x \;dx = (\tan x + C_1) - (x + C_2) = \tan x - x + (C_1 - C_2)$

Let $C = C_1 - C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{\tan x - x + C}$.

We need to find the integral of the function $\frac{2 - 3 \sin x}{\cos^2 x}$ with respect to $x$.

First, we can rewrite the integrand by splitting the fraction:

$\frac{2 - 3 \sin x}{\cos^2 x} = \frac{2}{\cos^2 x} - \frac{3 \sin x}{\cos^2 x}$

Using trigonometric identities, we know that $\frac{1}{\cos x} = \sec x$ and $\frac{\sin x}{\cos x} = \tan x$.

So, $\frac{2}{\cos^2 x} = 2 \cdot \frac{1}{\cos^2 x} = 2 \sec^2 x$.

And $\frac{3 \sin x}{\cos^2 x} = 3 \cdot \frac{\sin x}{\cos x \cdot \cos x} = 3 \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = 3 \tan x \sec x$.

The integrand becomes $2 \sec^2 x - 3 \sec x \tan x$.

Now, we need to find the integral of the simplified expression:

$\int (2 \sec^2 x - 3 \sec x \tan x) \;dx$

Using the linearity property of integration, we integrate each term separately:

$\int 2 \sec^2 x \;dx - \int 3 \sec x \tan x \;dx$

We can take the constants outside the integrals:

$2 \int \sec^2 x \;dx - 3 \int \sec x \tan x \;dx$

Using the standard integral formulas:

$\int \sec^2 x \;dx = \tan x + C_1$

$\int \sec x \tan x \;dx = \sec x + C_2$

Substituting these results back, we get:

$2(\tan x + C_1) - 3(\sec x + C_2)$

$2 \tan x + 2C_1 - 3 \sec x - 3C_2$

$2 \tan x - 3 \sec x + (2C_1 - 3C_2)$

Let $C = 2C_1 - 3C_2$ be the constant of integration.

Therefore, the integral is $\mathbf{2 \tan x - 3 \sec x + C}$.

To find the anti derivative of $\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$, we need to evaluate the integral:

$\int \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \;dx$

Rewrite the terms using exponents:

$\sqrt{x} = x^{1/2}$

$\frac{1}{\sqrt{x}} = x^{-1/2}$

So, the integral becomes:

$\int (x^{1/2} + x^{-1/2}) \;dx$

Using the linearity property of integrals, we integrate each term separately:

$\int x^{1/2} \;dx + \int x^{-1/2} \;dx$

Apply the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

For the first term ($n = 1/2$):

$\int x^{1/2} \;dx = \frac{x^{1/2 + 1}}{1/2 + 1} + C_1 = \frac{x^{3/2}}{3/2} + C_1 = \frac{2}{3}x^{3/2} + C_1$

For the second term ($n = -1/2$):

$\int x^{-1/2} \;dx = \frac{x^{-1/2 + 1}}{-1/2 + 1} + C_2 = \frac{x^{1/2}}{1/2} + C_2 = 2x^{1/2} + C_2$

Combine the results:

$\int (x^{1/2} + x^{-1/2}) \;dx = \frac{2}{3}x^{3/2} + 2x^{1/2} + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

The anti derivative is $\frac{2}{3}x^{3/2} + 2x^{1/2} + C$.

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\mathbf{\frac{2}{3} x^{\frac{3}{2}} + 2x^{\frac{1}{2}} + C}$.

Given:

$\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}$

$f(2) = 0$

To Find:

The function $f(x)$.

Solution:

Since $\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}$, the function $f(x)$ is the anti derivative (or integral) of $4x^3 - \frac{3}{x^4}$.

$f(x) = \int \left(4x^3 - \frac{3}{x^4}\right) \;dx$

Rewrite the term $\frac{3}{x^4}$ as $3x^{-4}$:

$f(x) = \int (4x^3 - 3x^{-4}) \;dx$

Using the linearity property of integration, we integrate term by term:

$f(x) = \int 4x^3 \;dx - \int 3x^{-4} \;dx$

$f(x) = 4 \int x^3 \;dx - 3 \int x^{-4} \;dx$

Apply the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$\int x^3 \;dx = \frac{x^{3+1}}{3+1} = \frac{x^4}{4}$

$\int x^{-4} \;dx = \frac{x^{-4+1}}{-4+1} = \frac{x^{-3}}{-3} = -\frac{1}{3}x^{-3}$

Substitute these results back into the expression for $f(x)$ and add the constant of integration $C$:

$f(x) = 4 \left(\frac{x^4}{4}\right) - 3 \left(-\frac{1}{3}x^{-3}\right) + C$

$f(x) = x^4 - (-x^{-3}) + C$

$f(x) = x^4 + x^{-3} + C$

$f(x) = x^4 + \frac{1}{x^3} + C$

Now, use the given condition $f(2) = 0$ to find the value of $C$. Substitute $x=2$ into the expression for $f(x)$:

$f(2) = (2)^4 + \frac{1}{(2)^3} + C$

$0 = 16 + \frac{1}{8} + C$

Solve for $C$:

$0 = \frac{16 \times 8}{8} + \frac{1}{8} + C$

$0 = \frac{128}{8} + \frac{1}{8} + C$

$0 = \frac{129}{8} + C$

$C = -\frac{129}{8}$

Substitute the value of $C$ back into the expression for $f(x)$:

$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$

Compare this result with the given options.

The function $f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$ matches option (A).

The correct answer is (A) $\mathbf{x^4 + \frac{1}{x^3} - \frac{129}{8}}$.

We will find the integrals using the method of substitution.

(i) $\int \sin mx \;dx$

Let $u = mx$. Then, $\frac{du}{dx} = m$, which means $dx = \frac{1}{m} du$.

Substituting this into the integral, we get:

$\int \sin u \cdot \frac{1}{m} \;du = \frac{1}{m} \int \sin u \;du$

We know that $\int \sin u \;du = -\cos u + C_1$.

So, $\frac{1}{m} \int \sin u \;du = \frac{1}{m} (-\cos u) + C = -\frac{1}{m} \cos u + C$.

Substitute back $u = mx$:

$-\frac{1}{m} \cos (mx) + C$

Therefore, $\int \sin mx \;dx = \mathbf{-\frac{1}{m} \cos (mx) + C}$.

(ii) $\int 2x \sin (x^2 + 1) \;dx$

Let $u = x^2 + 1$. Then, $\frac{du}{dx} = \frac{d}{dx}(x^2 + 1) = 2x$, which means $du = 2x \;dx$.

Substituting this into the integral, we get:

$\int \sin (x^2 + 1) \cdot (2x \;dx) = \int \sin u \;du$

We know that $\int \sin u \;du = -\cos u + C$.

Substitute back $u = x^2 + 1$:

$-\cos (x^2 + 1) + C$

Therefore, $\int 2x \sin (x^2 + 1) \;dx = \mathbf{-\cos (x^2 + 1) + C}$.

(iii) $\int \frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}} \;dx$

Let $u = \sqrt{x} = x^{1/2}$. Then, $\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

This means $du = \frac{1}{2\sqrt{x}} \;dx$, or $\frac{1}{\sqrt{x}} \;dx = 2 \;du$.

Substitute this into the integral:

$\int \tan^4 \sqrt{x} \sec^2 \sqrt{x} \cdot \frac{1}{\sqrt{x}} \;dx = \int \tan^4 u \sec^2 u \cdot (2 \;du) = 2 \int \tan^4 u \sec^2 u \;du$

Now, let $v = \tan u$. Then, $\frac{dv}{du} = \frac{d}{du}(\tan u) = \sec^2 u$, which means $dv = \sec^2 u \;du$.

Substitute this into the integral with respect to $u$:

$2 \int (\tan u)^4 (\sec^2 u \;du) = 2 \int v^4 \;dv$

Using the power rule $\int v^n \;dv = \frac{v^{n+1}}{n+1} + C_1$:

$2 \int v^4 \;dv = 2 \frac{v^{4+1}}{4+1} + C = 2 \frac{v^5}{5} + C = \frac{2}{5}v^5 + C$.

Substitute back $v = \tan u$:

$\frac{2}{5}(\tan u)^5 + C = \frac{2}{5} \tan^5 u + C$.

Substitute back $u = \sqrt{x}$:

$\frac{2}{5} \tan^5 \sqrt{x} + C$.

Therefore, $\int \frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}} \;dx = \mathbf{\frac{2}{5} \tan^5 \sqrt{x} + C}$.

(iv) $\int \frac{\sin (\tan^{−1} x)}{1 + x^2} \;dx$

Let $u = \tan^{-1} x$. Then, $\frac{du}{dx} = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}$.

This means $du = \frac{1}{1 + x^2} \;dx$.

Substitute this into the integral:

$\int \sin (\tan^{-1} x) \cdot \frac{1}{1 + x^2} \;dx = \int \sin u \;du$

We know that $\int \sin u \;du = -\cos u + C$.

Substitute back $u = \tan^{-1} x$:

$-\cos (\tan^{-1} x) + C$

We can simplify $\cos (\tan^{-1} x)$. Let $y = \tan^{-1} x$, so $\tan y = x$. Consider a right triangle where the opposite side is $x$ and the adjacent side is 1. The hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{x^2+1}$.

Then $\cos y = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{x^2+1}}$.

So, $\cos (\tan^{-1} x) = \frac{1}{\sqrt{x^2+1}}$.

The integral is $-\frac{1}{\sqrt{x^2+1}} + C$.

Therefore, $\int \frac{\sin (\tan^{−1} x)}{1 + x^2} \;dx = \mathbf{-\cos (\tan^{-1} x) + C}$ or $\mathbf{-\frac{1}{\sqrt{x^2+1}} + C}$.

We will find the integrals using appropriate techniques, mainly substitution and trigonometric identities.

(i) $\int \sin^3 x \cos^2 x \;dx$

We can rewrite $\sin^3 x$ as $\sin^2 x \sin x$. Use the identity $\sin^2 x = 1 - \cos^2 x$.

The integrand becomes $(1 - \cos^2 x) \cos^2 x \sin x = (\cos^2 x - \cos^4 x) \sin x$.

Let $u = \cos x$. Then $\frac{du}{dx} = -\sin x$, which means $du = -\sin x \;dx$, or $\sin x \;dx = -du$.

Substitute this into the integral:

$\int (\cos^2 x - \cos^4 x) \sin x \;dx = \int (u^2 - u^4) (-du)$

$\int -(u^2 - u^4) \;du = \int (u^4 - u^2) \;du$

Integrate term by term using the power rule:

$\int u^4 \;du - \int u^2 \;du = \frac{u^{4+1}}{4+1} - \frac{u^{2+1}}{2+1} + C = \frac{u^5}{5} - \frac{u^3}{3} + C$.

Substitute back $u = \cos x$:

$\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$.

Therefore, $\int \sin^3 x \cos^2 x \;dx = \mathbf{\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C}$.

(ii) $\int \frac{\sin x}{\sin (x + a)} \;dx$

Let $u = x + a$. Then $du = dx$. Also, $x = u - a$.

Substitute this into the integral:

$\int \frac{\sin (u - a)}{\sin u} \;du$

Use the trigonometric identity for $\sin(A-B) = \sin A \cos B - \cos A \sin B$:

$\sin(u-a) = \sin u \cos a - \cos u \sin a$.

The integrand becomes $\frac{\sin u \cos a - \cos u \sin a}{\sin u} = \frac{\sin u \cos a}{\sin u} - \frac{\cos u \sin a}{\sin u} = \cos a - \cot u \sin a$.

Now, integrate with respect to $u$:

$\int (\cos a - \cot u \sin a) \;du = \int \cos a \;du - \int \cot u \sin a \;du$

Since $\cos a$ and $\sin a$ are constants with respect to $u$:

$\int \cos a \;du = \cos a \int 1 \;du = u \cos a + C_1$

$\int \cot u \sin a \;du = \sin a \int \cot u \;du = \sin a \ln |\sin u| + C_2$

So, the integral is $u \cos a - \sin a \ln |\sin u| + C$.

Substitute back $u = x + a$:

$(x + a) \cos a - \sin a \ln |\sin (x + a)| + C$.

Therefore, $\int \frac{\sin x}{\sin (x + a)} \;dx = \mathbf{(x + a) \cos a - \sin a \ln |\sin (x + a)| + C}$.

(iii) $\int \frac{1}{1 + \tan x} \;dx$

Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$:

$\frac{1}{1 + \frac{\sin x}{\cos x}} = \frac{1}{\frac{\cos x + \sin x}{\cos x}} = \frac{\cos x}{\cos x + \sin x}$.

The integral is $\int \frac{\cos x}{\cos x + \sin x} \;dx$.

This is a common type of integral. We can use a trick: write the numerator in terms of the denominator and its derivative.

Let the numerator be $A(\cos x + \sin x) + B \frac{d}{dx}(\cos x + \sin x)$.

$\frac{d}{dx}(\cos x + \sin x) = -\sin x + \cos x$.

So, $\cos x = A(\cos x + \sin x) + B (-\sin x + \cos x)$

$\cos x = A \cos x + A \sin x - B \sin x + B \cos x$

$\cos x = (A + B) \cos x + (A - B) \sin x$.

Comparing coefficients of $\cos x$ and $\sin x$ on both sides:

Adding (1) and (2): $2A = 1 \implies A = \frac{1}{2}$.

Substituting $A = \frac{1}{2}$ into (2): $\frac{1}{2} - B = 0 \implies B = \frac{1}{2}$.

So, $\cos x = \frac{1}{2}(\cos x + \sin x) + \frac{1}{2}(\cos x - \sin x)$.

The integrand becomes $\frac{\frac{1}{2}(\cos x + \sin x) + \frac{1}{2}(\cos x - \sin x)}{\cos x + \sin x} = \frac{1}{2} \frac{\cos x + \sin x}{\cos x + \sin x} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x}$.

For $\cos x + \sin x \neq 0$, this simplifies to $\frac{1}{2} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x}$.

Now, integrate this expression:

$\int \left(\frac{1}{2} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x}\right) \;dx = \int \frac{1}{2} \;dx + \int \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x} \;dx$

$\int \frac{1}{2} \;dx = \frac{1}{2}x + C_1$.

For the second integral, let $v = \cos x + \sin x$. Then $\frac{dv}{dx} = -\sin x + \cos x$, so $dv = (\cos x - \sin x) \;dx$.

The second integral becomes $\int \frac{1}{2} \frac{dv}{v} = \frac{1}{2} \int \frac{1}{v} \;dv$.

$\frac{1}{2} \int \frac{1}{v} \;dv = \frac{1}{2} \ln |v| + C_2$.

Substitute back $v = \cos x + \sin x$:

$\frac{1}{2} \ln |\cos x + \sin x| + C_2$.

Combining the results:

$\int \frac{1}{1 + \tan x} \;dx = \frac{1}{2}x + \frac{1}{2} \ln |\cos x + \sin x| + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, $\int \frac{1}{1 + \tan x} \;dx = \mathbf{\frac{1}{2}x + \frac{1}{2} \ln |\cos x + \sin x| + C}$.

We need to find the integral of the function $\frac{2x}{1 + x^2}$ with respect to $x$.

Let the integral be $I = \int \frac{2x}{1 + x^2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 + x^2$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + x^2) = 0 + 2x = 2x$

This gives us the differential $du = 2x \;dx$.

Now substitute $u = 1 + x^2$ and $du = 2x \;dx$ into the integral:

$I = \int \frac{1}{1 + x^2} \cdot (2x \;dx)$

$I = \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C$

where $C$ is the constant of integration.

Substitute back $u = 1 + x^2$ to express the result in terms of $x$:

$I = \ln |1 + x^2| + C$

Since $1 + x^2 \geq 1$ for all real values of $x$, $1 + x^2$ is always positive. Therefore, $|1 + x^2| = 1 + x^2$, and the absolute value is not strictly necessary.

$I = \ln (1 + x^2) + C$

Therefore, the integral of $\frac{2x}{1 + x^2}$ is $\mathbf{\ln (1 + x^2) + C}$.

We need to find the integral of the function $\frac{(\log x)^2}{x}$ with respect to $x$.

Let the integral be $I = \int \frac{(\log x)^2}{x} \;dx$.

We can solve this integral using the method of substitution.

Let $u = \log x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

This gives us the differential $du = \frac{1}{x} \;dx$.

Now substitute $u = \log x$ and $du = \frac{1}{x} \;dx$ into the integral:

$I = \int (\log x)^2 \cdot \left(\frac{1}{x} \;dx\right)$

$I = \int u^2 \;du$

The integral of $u^2$ with respect to $u$ is a standard integral using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C$:

$\int u^2 \;du = \frac{u^{2+1}}{2+1} + C = \frac{u^3}{3} + C$

where $C$ is the constant of integration.

Substitute back $u = \log x$ to express the result in terms of $x$:

$I = \frac{(\log x)^3}{3} + C$

Therefore, the integral of $\frac{(\log x)^2}{x}$ is $\mathbf{\frac{(\log x)^3}{3} + C}$.

We need to find the integral of the function $\frac{1}{x + x \log x}$ with respect to $x$.

Let the integral be $I = \int \frac{1}{x + x \log x} \;dx$.

First, simplify the integrand by factoring out $x$ from the denominator:

$\frac{1}{x + x \log x} = \frac{1}{x(1 + \log x)}$

Now, the integral is $\int \frac{1}{x(1 + \log x)} \;dx$. We can rewrite this as $\int \frac{1}{1 + \log x} \cdot \frac{1}{x} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 + \log x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + \log x) = 0 + \frac{1}{x} = \frac{1}{x}$

This gives us the differential $du = \frac{1}{x} \;dx$.

Now substitute $u = 1 + \log x$ and $du = \frac{1}{x} \;dx$ into the integral:

$I = \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C$

where $C$ is the constant of integration.

Substitute back $u = 1 + \log x$ to express the result in terms of $x$:

$I = \ln |1 + \log x| + C$

Therefore, the integral of $\frac{1}{x + x \log x}$ is $\mathbf{\ln |1 + \log x| + C}$.

We need to find the integral of the function $\sin x \sin (\cos x)$ with respect to $x$.

Let the integral be $I = \int \sin x \sin (\cos x) \;dx$.

We can solve this integral using the method of substitution.

Let $u = \cos x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x$

This gives us the differential $du = -\sin x \;dx$, or $\sin x \;dx = -du$.

Now substitute $u = \cos x$ and $\sin x \;dx = -du$ into the integral:

$I = \int \sin (\cos x) \cdot (\sin x \;dx)$

$I = \int \sin u \cdot (-du)$

$I = - \int \sin u \;du$

The integral of $\sin u$ with respect to $u$ is a standard integral:

$\int \sin u \;du = -\cos u + C_1$

where $C_1$ is the constant of integration.

So, $I = - (-\cos u + C_1) = \cos u - C_1$.

Let $C = -C_1$ be the constant of integration.

$I = \cos u + C$

Substitute back $u = \cos x$ to express the result in terms of $x$:

$I = \cos (\cos x) + C$

Therefore, the integral of $\sin x \sin (\cos x)$ is $\mathbf{\cos (\cos x) + C}$.

We need to find the integral of the function $\sin(ax + b) \cos(ax + b)$ with respect to $x$.

Let the integral be $I = \int \sin(ax + b) \cos(ax + b) \;dx$.

We can solve this integral using the method of substitution.

Let $u = \sin(ax + b)$.

Differentiate $u$ with respect to $x$ using the chain rule:

$\frac{du}{dx} = \frac{d}{dx}(\sin(ax + b)) = \cos(ax + b) \cdot \frac{d}{dx}(ax + b)$

$\frac{du}{dx} = \cos(ax + b) \cdot a = a \cos(ax + b)$

This gives us the differential $du = a \cos(ax + b) \;dx$.

From this, we can express $\cos(ax + b) \;dx$ as $\frac{1}{a} du$ (assuming $a \neq 0$).

Now substitute $u = \sin(ax + b)$ and $\cos(ax + b) \;dx = \frac{1}{a} du$ into the integral:

$I = \int u \cdot \left(\frac{1}{a} du\right)$

$I = \frac{1}{a} \int u \;du$

The integral of $u$ with respect to $u$ is a standard integral using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u \;du = \frac{u^{1+1}}{1+1} + C' = \frac{u^2}{2} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{a} \left(\frac{u^2}{2} + C'\right) = \frac{u^2}{2a} + \frac{C'}{a}$

Let $C = \frac{C'}{a}$ be the new constant of integration.

$I = \frac{u^2}{2a} + C$

Finally, substitute back $u = \sin(ax + b)$ to express the result in terms of $x$:

$I = \frac{(\sin(ax + b))^2}{2a} + C = \frac{\sin^2(ax + b)}{2a} + C$

Alternatively, we could use the substitution $v = \cos(ax+b)$, which would give $dv = -a \sin(ax+b) dx$. Then $\sin(ax+b) dx = -\frac{1}{a} dv$. The integral becomes $\int v (-\frac{1}{a} dv) = -\frac{1}{a} \int v dv = -\frac{1}{a} \frac{v^2}{2} + C' = -\frac{\cos^2(ax+b)}{2a} + C$. This is also a valid anti derivative, differing by a constant from the first result due to $\sin^2\theta + \cos^2\theta = 1$.

Therefore, the integral of $\sin(ax + b) \cos(ax + b)$ is $\mathbf{\frac{\sin^2(ax + b)}{2a} + C}$.

We need to find the integral of the function $\sqrt{ax + b}$ with respect to $x$.

Let the integral be $I = \int \sqrt{ax + b} \;dx$.

Rewrite the square root as a power: $\sqrt{ax + b} = (ax + b)^{1/2}$.

$I = \int (ax + b)^{1/2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = ax + b$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(ax + b) = a$

This gives us the differential $du = a \;dx$.

From this, we can express $dx$ as $\frac{1}{a} du$ (assuming $a \neq 0$).

Now substitute $u = ax + b$ and $dx = \frac{1}{a} du$ into the integral:

$I = \int u^{1/2} \cdot \left(\frac{1}{a} du\right)$

$I = \frac{1}{a} \int u^{1/2} \;du$

The integral of $u^{1/2}$ with respect to $u$ is a standard integral using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C' = \frac{u^{3/2}}{3/2} + C' = \frac{2}{3}u^{3/2} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{a} \left(\frac{2}{3}u^{3/2} + C'\right) = \frac{2}{3a}u^{3/2} + \frac{C'}{a}$

Let $C = \frac{C'}{a}$ be the new constant of integration.

$I = \frac{2}{3a}u^{3/2} + C$

Finally, substitute back $u = ax + b$ to express the result in terms of $x$:

$I = \frac{2}{3a}(ax + b)^{3/2} + C$

Therefore, the integral of $\sqrt{ax + b}$ is $\mathbf{\frac{2}{3a}(ax + b)^{\frac{3}{2}} + C}$.

We need to find the integral of the function $x \sqrt{x + 2}$ with respect to $x$.

Let the integral be $I = \int x \sqrt{x + 2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = x + 2$.

From this substitution, we have $x = u - 2$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x + 2) = 1$

This gives us the differential $du = dx$.

Now substitute $u = x + 2$, $x = u - 2$, and $dx = du$ into the integral:

$I = \int (u - 2) \sqrt{u} \;du$

$I = \int (u - 2) u^{1/2} \;du$

Distribute $u^{1/2}$ through the parenthesis:

$I = \int (u \cdot u^{1/2} - 2 \cdot u^{1/2}) \;du$

Using the exponent rule $u^m \cdot u^n = u^{m+n}$:

$u \cdot u^{1/2} = u^1 \cdot u^{1/2} = u^{1 + 1/2} = u^{3/2}$

So, the integrand is $u^{3/2} - 2u^{1/2}$.

$I = \int (u^{3/2} - 2u^{1/2}) \;du$

Using the linearity property of integration, we integrate each term separately:

$I = \int u^{3/2} \;du - \int 2u^{1/2} \;du$

$I = \int u^{3/2} \;du - 2 \int u^{1/2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u^{3/2} \;du = \frac{u^{3/2 + 1}}{3/2 + 1} + C_1 = \frac{u^{5/2}}{5/2} + C_1 = \frac{2}{5}u^{5/2} + C_1$

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C_2 = \frac{u^{3/2}}{3/2} + C_2 = \frac{2}{3}u^{3/2} + C_2$

Substitute these results back into the expression for $I$:

$I = \left(\frac{2}{5}u^{5/2} + C_1\right) - 2 \left(\frac{2}{3}u^{3/2} + C_2\right)$

$I = \frac{2}{5}u^{5/2} + C_1 - \frac{4}{3}u^{3/2} - 2C_2$

$I = \frac{2}{5}u^{5/2} - \frac{4}{3}u^{3/2} + (C_1 - 2C_2)$

Let $C = C_1 - 2C_2$ be the constant of integration.

$I = \frac{2}{5}u^{5/2} - \frac{4}{3}u^{3/2} + C$

Finally, substitute back $u = x + 2$ to express the result in terms of $x$:

$I = \frac{2}{5}(x + 2)^{5/2} - \frac{4}{3}(x + 2)^{3/2} + C$

Therefore, the integral of $x \sqrt{x + 2}$ is $\mathbf{\frac{2}{5}(x + 2)^{\frac{5}{2}} - \frac{4}{3}(x + 2)^{\frac{3}{2}} + C}$.

We need to find the integral of the function $x \sqrt{1 + 2x^2}$ with respect to $x$.

Let the integral be $I = \int x \sqrt{1 + 2x^2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 + 2x^2$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + 2x^2) = 0 + 2(2x) = 4x$

This gives us the differential $du = 4x \;dx$.

From this, we can express $x \;dx$ as $\frac{1}{4} du$.

Now substitute $u = 1 + 2x^2$ and $x \;dx = \frac{1}{4} du$ into the integral:

$I = \int \sqrt{1 + 2x^2} \cdot (x \;dx)$

$I = \int \sqrt{u} \cdot \left(\frac{1}{4} du\right)$

$I = \frac{1}{4} \int \sqrt{u} \;du$

Rewrite the square root as a power: $\sqrt{u} = u^{1/2}$.

$I = \frac{1}{4} \int u^{1/2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C' = \frac{u^{3/2}}{3/2} + C' = \frac{2}{3}u^{3/2} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{4} \left(\frac{2}{3}u^{3/2} + C'\right) = \frac{2}{12}u^{3/2} + \frac{C'}{4} = \frac{1}{6}u^{3/2} + \frac{C'}{4}$

Let $C = \frac{C'}{4}$ be the new constant of integration.

$I = \frac{1}{6}u^{3/2} + C$

Finally, substitute back $u = 1 + 2x^2$ to express the result in terms of $x$:

$I = \frac{1}{6}(1 + 2x^2)^{3/2} + C$

Therefore, the integral of $x \sqrt{1 + 2x^2}$ is $\mathbf{\frac{1}{6}(1 + 2x^2)^{\frac{3}{2}} + C}$.

We need to find the integral of the function $(4x + 2) \sqrt{x^2 + x + 1}$ with respect to $x$.

Let the integral be $I = \int (4x + 2) \sqrt{x^2 + x + 1} \;dx$.

We can factor out a 2 from the term $(4x + 2)$: $4x + 2 = 2(2x + 1)$.

The integral becomes $I = \int 2(2x + 1) \sqrt{x^2 + x + 1} \;dx = 2 \int (2x + 1) \sqrt{x^2 + x + 1} \;dx$.

We can solve this integral using the method of substitution.

Let $u = x^2 + x + 1$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^2 + x + 1) = 2x + 1 + 0 = 2x + 1$

This gives us the differential $du = (2x + 1) \;dx$.

Now substitute $u = x^2 + x + 1$ and $du = (2x + 1) \;dx$ into the integral:

$I = 2 \int \sqrt{x^2 + x + 1} \cdot (2x + 1) \;dx$

$I = 2 \int \sqrt{u} \;du$

Rewrite the square root as a power: $\sqrt{u} = u^{1/2}$.

$I = 2 \int u^{1/2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C' = \frac{u^{3/2}}{3/2} + C' = \frac{2}{3}u^{3/2} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = 2 \left(\frac{2}{3}u^{3/2} + C'\right) = \frac{4}{3}u^{3/2} + 2C'$

Let $C = 2C'$ be the new constant of integration.

$I = \frac{4}{3}u^{3/2} + C$

Finally, substitute back $u = x^2 + x + 1$ to express the result in terms of $x$:

$I = \frac{4}{3}(x^2 + x + 1)^{3/2} + C$

Therefore, the integral of $(4x + 2) \sqrt{x^2 + x + 1}$ is $\mathbf{\frac{4}{3}(x^2 + x + 1)^{\frac{3}{2}} + C}$.

We need to find the integral of the function $\frac{1}{x − \sqrt{x}}$ with respect to $x$.

Let the integral be $I = \int \frac{1}{x − \sqrt{x}} \;dx$.

First, simplify the denominator by factoring out $\sqrt{x}$:

$x - \sqrt{x} = \sqrt{x} \cdot \sqrt{x} - \sqrt{x} = \sqrt{x}(\sqrt{x} - 1)$

The integrand becomes $\frac{1}{\sqrt{x}(\sqrt{x} - 1)}$.

So, the integral is $I = \int \frac{1}{\sqrt{x}(\sqrt{x} - 1)} \;dx$. We can rewrite this as $\int \frac{1}{\sqrt{x} - 1} \cdot \frac{1}{\sqrt{x}} \;dx$.

We can solve this integral using the method of substitution.

Let $u = \sqrt{x} - 1$.

Differentiate $u$ with respect to $x$. Rewrite $\sqrt{x}$ as $x^{1/2}$:

$u = x^{1/2} - 1$

$\frac{du}{dx} = \frac{d}{dx}(x^{1/2} - 1) = \frac{1}{2}x^{1/2 - 1} - 0 = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

This gives us the differential $du = \frac{1}{2\sqrt{x}} \;dx$.

From this, we can express $\frac{1}{\sqrt{x}} \;dx$ as $2 \;du$.

Now substitute $u = \sqrt{x} - 1$ and $\frac{1}{\sqrt{x}} \;dx = 2 \;du$ into the integral:

$I = \int \frac{1}{\sqrt{x} - 1} \cdot \left(\frac{1}{\sqrt{x}} \;dx\right)$

$I = \int \frac{1}{u} \cdot (2 \;du)$

$I = 2 \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = 2 (\ln |u| + C') = 2 \ln |u| + 2C'$

Let $C = 2C'$ be the new constant of integration.

$I = 2 \ln |u| + C$

Finally, substitute back $u = \sqrt{x} - 1$ to express the result in terms of $x$:

$I = 2 \ln |\sqrt{x} - 1| + C$

Therefore, the integral of $\frac{1}{x − \sqrt{x}}$ is $\mathbf{2 \ln |\sqrt{x} - 1| + C}$.

We need to find the integral of the function $\frac{x}{\sqrt{x + 4}}$ with respect to $x$ for $x > 0$.

Let the integral be $I = \int \frac{x}{\sqrt{x + 4}} \;dx$.

We can solve this integral using the method of substitution.

Let $u = x + 4$.

From this substitution, we can express $x$ in terms of $u$: $x = u - 4$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x + 4) = 1$

This gives us the differential $du = dx$.

Now substitute $u = x + 4$, $x = u - 4$, and $dx = du$ into the integral:

$I = \int \frac{u - 4}{\sqrt{u}} \;du$

Rewrite $\sqrt{u}$ as $u^{1/2}$.

$I = \int \frac{u - 4}{u^{1/2}} \;du$

Simplify the integrand by dividing each term in the numerator by the denominator:

$\frac{u - 4}{u^{1/2}} = \frac{u}{u^{1/2}} - \frac{4}{u^{1/2}}$

Using the exponent rule $\frac{u^m}{u^n} = u^{m-n}$:

$\frac{u}{u^{1/2}} = u^{1 - 1/2} = u^{1/2}$

$\frac{4}{u^{1/2}} = 4u^{-1/2}$

So, the integral becomes:

$I = \int (u^{1/2} - 4u^{-1/2}) \;du$

Using the linearity property of integration, we integrate each term separately:

$I = \int u^{1/2} \;du - \int 4u^{-1/2} \;du$

$I = \int u^{1/2} \;du - 4 \int u^{-1/2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$):

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C_1 = \frac{u^{3/2}}{3/2} + C_1 = \frac{2}{3}u^{3/2} + C_1$

$\int u^{-1/2} \;du = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_2 = \frac{u^{1/2}}{1/2} + C_2 = 2u^{1/2} + C_2$

Substitute these results back into the expression for $I$:

$I = \left(\frac{2}{3}u^{3/2} + C_1\right) - 4 \left(2u^{1/2} + C_2\right)$

$I = \frac{2}{3}u^{3/2} + C_1 - 8u^{1/2} - 8C_2$

$I = \frac{2}{3}u^{3/2} - 8u^{1/2} + (C_1 - 8C_2)$

Let $C = C_1 - 8C_2$ be the constant of integration.

$I = \frac{2}{3}u^{3/2} - 8u^{1/2} + C$

Finally, substitute back $u = x + 4$ to express the result in terms of $x$:

$I = \frac{2}{3}(x + 4)^{3/2} - 8(x + 4)^{1/2} + C$

$I = \frac{2}{3}(x + 4)\sqrt{x + 4} - 8\sqrt{x + 4} + C$

We can factor out $\sqrt{x+4}$:

$I = \sqrt{x+4} \left( \frac{2}{3}(x+4) - 8 \right) + C$

$I = \sqrt{x+4} \left( \frac{2}{3}x + \frac{8}{3} - \frac{24}{3} \right) + C$

$I = \sqrt{x+4} \left( \frac{2}{3}x - \frac{16}{3} \right) + C$

$I = \frac{2}{3}(x - 8)\sqrt{x + 4} + C$

Therefore, the integral of $\frac{x}{\sqrt{x + 4}}$ is $\mathbf{\frac{2}{3}(x + 4)^{\frac{3}{2}} - 8(x + 4)^{\frac{1}{2}} + C}$ or $\mathbf{\frac{2}{3}(x - 8)\sqrt{x + 4} + C}$.

We need to find the integral of the function $(x^3 − 1)^{\frac{1}{3}} x^5$ with respect to $x$.

Let the integral be $I = \int (x^3 − 1)^{\frac{1}{3}} x^5 \;dx$.

We can rewrite $x^5$ as $x^3 \cdot x^2$. This will be useful for substitution.

$I = \int (x^3 − 1)^{\frac{1}{3}} x^3 \cdot x^2 \;dx$.

We can solve this integral using the method of substitution.

Let $u = x^3 - 1$.

From this substitution, we can express $x^3$ in terms of $u$: $x^3 = u + 1$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^3 - 1) = 3x^2$

This gives us the differential $du = 3x^2 \;dx$.

From this, we can express $x^2 \;dx$ as $\frac{1}{3} du$.

Now substitute $u = x^3 - 1$, $x^3 = u + 1$, and $x^2 \;dx = \frac{1}{3} du$ into the integral:

$I = \int (x^3 − 1)^{\frac{1}{3}} x^3 \cdot (x^2 \;dx)$

$I = \int u^{\frac{1}{3}} (u + 1) \cdot \left(\frac{1}{3} du\right)$

$I = \frac{1}{3} \int u^{1/3} (u + 1) \;du$

Distribute $u^{1/3}$ through the parenthesis:

$u^{1/3} (u + 1) = u^{1/3} \cdot u^1 + u^{1/3} \cdot 1$

Using the exponent rule $u^m \cdot u^n = u^{m+n}$:

$u^{1/3} \cdot u^1 = u^{1/3 + 1} = u^{1/3 + 3/3} = u^{4/3}$

So, the integrand is $u^{4/3} + u^{1/3}$.

$I = \frac{1}{3} \int (u^{4/3} + u^{1/3}) \;du$

Using the linearity property of integration, we integrate each term separately:

$I = \frac{1}{3} \left( \int u^{4/3} \;du + \int u^{1/3} \;du \right)$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$):

$\int u^{4/3} \;du = \frac{u^{4/3 + 1}}{4/3 + 1} + C_1 = \frac{u^{7/3}}{7/3} + C_1 = \frac{3}{7}u^{7/3} + C_1$

$\int u^{1/3} \;du = \frac{u^{1/3 + 1}}{1/3 + 1} + C_2 = \frac{u^{4/3}}{4/3} + C_2 = \frac{3}{4}u^{4/3} + C_2$

Substitute these results back into the expression for $I$:

$I = \frac{1}{3} \left( \frac{3}{7}u^{7/3} + C_1 + \frac{3}{4}u^{4/3} + C_2 \right)$

$I = \frac{1}{3} \left( \frac{3}{7}u^{7/3} + \frac{3}{4}u^{4/3} \right) + \frac{1}{3}(C_1 + C_2)$

$I = \frac{1}{7}u^{7/3} + \frac{1}{4}u^{4/3} + C$

Let $C = \frac{1}{3}(C_1 + C_2)$ be the constant of integration.

Finally, substitute back $u = x^3 - 1$ to express the result in terms of $x$:

$I = \frac{1}{7}(x^3 - 1)^{7/3} + \frac{1}{4}(x^3 - 1)^{4/3} + C$

Therefore, the integral of $(x^3 − 1)^{\frac{1}{3}} x^5$ is $\mathbf{\frac{1}{7}(x^3 - 1)^{\frac{7}{3}} + \frac{1}{4}(x^3 - 1)^{\frac{4}{3}} + C}$.

We need to find the integral of the function $\frac{x^2}{(2 + 3x^3)^3}$ with respect to $x$.

Let the integral be $I = \int \frac{x^2}{(2 + 3x^3)^3} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 2 + 3x^3$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(2 + 3x^3) = 0 + 3(3x^{3-1}) = 9x^2$

This gives us the differential $du = 9x^2 \;dx$.

From this, we can express $x^2 \;dx$ as $\frac{1}{9} du$.

Now substitute $u = 2 + 3x^3$ and $x^2 \;dx = \frac{1}{9} du$ into the integral:

$I = \int \frac{1}{(2 + 3x^3)^3} \cdot (x^2 \;dx)$

$I = \int \frac{1}{u^3} \cdot \left(\frac{1}{9} du\right)$

$I = \frac{1}{9} \int \frac{1}{u^3} \;du$

Rewrite $\frac{1}{u^3}$ using negative exponents: $\frac{1}{u^3} = u^{-3}$.

$I = \frac{1}{9} \int u^{-3} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$). Here $n = -3$:

$\int u^{-3} \;du = \frac{u^{-3+1}}{-3+1} + C' = \frac{u^{-2}}{-2} + C' = -\frac{1}{2}u^{-2} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{9} \left(-\frac{1}{2}u^{-2} + C'\right)$

$I = -\frac{1}{18}u^{-2} + \frac{C'}{9}$

Rewrite $u^{-2}$ as $\frac{1}{u^2}$:

$I = -\frac{1}{18u^2} + \frac{C'}{9}$

Let $C = \frac{C'}{9}$ be the new constant of integration.

$I = -\frac{1}{18u^2} + C$

Finally, substitute back $u = 2 + 3x^3$ to express the result in terms of $x$:

$I = -\frac{1}{18(2 + 3x^3)^2} + C$

Therefore, the integral of $\frac{x^2}{(2 + 3x^3)^3}$ is $\mathbf{-\frac{1}{18(2 + 3x^3)^2} + C}$.

We need to find the integral of the function $\frac{1}{x(\log x)^m}$ with respect to $x$, where $x > 0$ and $m \neq 1$.

Let the integral be $I = \int \frac{1}{x(\log x)^m} \;dx$.

We can rewrite the integrand as $\frac{1}{(\log x)^m} \cdot \frac{1}{x}$.

$I = \int \frac{1}{(\log x)^m} \cdot \frac{1}{x} \;dx$.

We can solve this integral using the method of substitution.

Let $u = \log x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

This gives us the differential $du = \frac{1}{x} \;dx$.

Now substitute $u = \log x$ and $du = \frac{1}{x} \;dx$ into the integral:

$I = \int \frac{1}{u^m} \;du$

Rewrite $\frac{1}{u^m}$ using negative exponents: $\frac{1}{u^m} = u^{-m}$.

$I = \int u^{-m} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$). Here $n = -m$. Since $m \neq 1$, $-m \neq -1$, so we can use the power rule:

$\int u^{-m} \;du = \frac{u^{-m+1}}{-m+1} + C' = \frac{u^{1-m}}{1-m} + C'$

where $C'$ is the constant of integration.

Let $C = C'$ be the new constant of integration.

$I = \frac{u^{1-m}}{1-m} + C$

Finally, substitute back $u = \log x$ to express the result in terms of $x$:

$I = \frac{(\log x)^{1-m}}{1-m} + C$

Therefore, the integral of $\frac{1}{x(\log x)^m}$ is $\mathbf{\frac{(\log x)^{1-m}}{1 - m} + C}$.

We need to find the integral of the function $\frac{x}{9 − 4x^2}$ with respect to $x$.

Let the integral be $I = \int \frac{x}{9 − 4x^2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 9 - 4x^2$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(9 - 4x^2) = 0 - 4(2x) = -8x$

This gives us the differential $du = -8x \;dx$.

From this, we can express $x \;dx$ as $-\frac{1}{8} du$.

Now substitute $u = 9 - 4x^2$ and $x \;dx = -\frac{1}{8} du$ into the integral:

$I = \int \frac{1}{9 − 4x^2} \cdot (x \;dx)$

$I = \int \frac{1}{u} \cdot \left(-\frac{1}{8} du\right)$

$I = -\frac{1}{8} \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = -\frac{1}{8} (\ln |u| + C') = -\frac{1}{8} \ln |u| - \frac{C'}{8}$

Let $C = -\frac{C'}{8}$ be the new constant of integration.

$I = -\frac{1}{8} \ln |u| + C$

Finally, substitute back $u = 9 - 4x^2$ to express the result in terms of $x$:

$I = -\frac{1}{8} \ln |9 - 4x^2| + C$

Therefore, the integral of $\frac{x}{9 − 4x^2}$ is $\mathbf{-\frac{1}{8} \ln |9 - 4x^2| + C}$.

We need to find the integral of the function $e^{2x+3}$ with respect to $x$.

Let the integral be $I = \int e^{2x+3} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 2x + 3$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(2x + 3) = 2$

This gives us the differential $du = 2 \;dx$.

From this, we can express $dx$ as $\frac{1}{2} du$.

Now substitute $u = 2x + 3$ and $dx = \frac{1}{2} du$ into the integral:

$I = \int e^{u} \cdot \left(\frac{1}{2} du\right)$

$I = \frac{1}{2} \int e^{u} \;du$

The integral of $e^u$ with respect to $u$ is a standard integral:

$\int e^u \;du = e^u + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{2} (e^u + C') = \frac{1}{2}e^u + \frac{C'}{2}$

Let $C = \frac{C'}{2}$ be the new constant of integration.

$I = \frac{1}{2}e^u + C$

Finally, substitute back $u = 2x + 3$ to express the result in terms of $x$:

$I = \frac{1}{2}e^{2x+3} + C$

Therefore, the integral of $e^{2x+3}$ is $\mathbf{\frac{1}{2}e^{2x+3} + C}$.

We need to find the integral of the function $\frac{x}{e^{x^2}}$ with respect to $x$.

Let the integral be $I = \int \frac{x}{e^{x^2}} \;dx$.

Rewrite the integrand using negative exponents: $\frac{x}{e^{x^2}} = x e^{-x^2}$.

$I = \int x e^{-x^2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = -x^2$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(-x^2) = -2x$

This gives us the differential $du = -2x \;dx$.

From this, we can express $x \;dx$ as $-\frac{1}{2} du$.

Now substitute $u = -x^2$ and $x \;dx = -\frac{1}{2} du$ into the integral:

$I = \int e^{-x^2} \cdot (x \;dx)$

$I = \int e^{u} \cdot \left(-\frac{1}{2} du\right)$

$I = -\frac{1}{2} \int e^{u} \;du$

The integral of $e^u$ with respect to $u$ is a standard integral:

$\int e^u \;du = e^u + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = -\frac{1}{2} (e^u + C') = -\frac{1}{2}e^u - \frac{C'}{2}$

Let $C = - \frac{C'}{2}$ be the new constant of integration.

$I = -\frac{1}{2}e^u + C$

Finally, substitute back $u = -x^2$ to express the result in terms of $x$:

$I = -\frac{1}{2}e^{-x^2} + C$

Therefore, the integral of $\frac{x}{e^{x^2}}$ is $\mathbf{-\frac{1}{2}e^{-x^2} + C}$.

We need to find the integral of the function $\frac{e^{\tan^{-1} x}}{1 + x^2}$ with respect to $x$.

Let the integral be $I = \int \frac{e^{\tan^{-1} x}}{1 + x^2} \;dx$.

We can rewrite the integrand as $e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2}$.

$I = \int e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = \tan^{-1} x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}$

This gives us the differential $du = \frac{1}{1 + x^2} \;dx$.

Now substitute $u = \tan^{-1} x$ and $du = \frac{1}{1 + x^2} \;dx$ into the integral:

$I = \int e^{u} \;du$

The integral of $e^u$ with respect to $u$ is a standard integral:

$\int e^u \;du = e^u + C$

where $C$ is the constant of integration.

Substitute back $u = \tan^{-1} x$ to express the result in terms of $x$:

$I = e^{\tan^{-1} x} + C$

Therefore, the integral of $\frac{e^{\tan^{-1} x}}{1 + x^2}$ is $\mathbf{e^{\tan^{-1} x} + C}$.

We need to find the integral of the function $\frac{e^{2x} − 1}{e^{2x} + 1}$ with respect to $x$.

Let the integral be $I = \int \frac{e^{2x} − 1}{e^{2x} + 1} \;dx$.

We can simplify the integrand by dividing the numerator and the denominator by $e^x$:

$\frac{e^{2x} - 1}{e^{2x} + 1} = \frac{(e^{2x} - 1)/e^x}{(e^{2x} + 1)/e^x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

So, the integral becomes $I = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} \;dx$.

We can solve this integral using the method of substitution.

Let $u = e^x + e^{-x}$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(e^x + e^{-x})$

$\frac{du}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{-x})$

$\frac{du}{dx} = e^x + e^{-x} \cdot \frac{d}{dx}(-x)$

$\frac{du}{dx} = e^x + e^{-x} \cdot (-1)$

$\frac{du}{dx} = e^x - e^{-x}$

This gives us the differential $du = (e^x - e^{-x}) \;dx$.

Now substitute $u = e^x + e^{-x}$ and $du = (e^x - e^{-x}) \;dx$ into the integral:

$I = \int \frac{1}{e^x + e^{-x}} \cdot (e^x - e^{-x}) \;dx$

$I = \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C$

where $C$ is the constant of integration.

Substitute back $u = e^x + e^{-x}$ to express the result in terms of $x$:

$I = \ln |e^x + e^{-x}| + C$

Since $e^x > 0$ and $e^{-x} > 0$ for all real values of $x$, their sum $e^x + e^{-x}$ is always positive. Thus, $|e^x + e^{-x}| = e^x + e^{-x}$.

$I = \ln (e^x + e^{-x}) + C$

Therefore, the integral of $\frac{e^{2x} − 1}{e^{2x} + 1}$ is $\mathbf{\ln (e^x + e^{-x}) + C}$.

We need to find the integral of the function $\frac{e^{2x} − e^{−2x}}{e^{2x} + e^{−2x}}$ with respect to $x$.

Let the integral be $I = \int \frac{e^{2x} − e^{−2x}}{e^{2x} + e^{−2x}} \;dx$.

We can solve this integral using the method of substitution.

Let $u = e^{2x} + e^{-2x}$.

Differentiate $u$ with respect to $x$ using the chain rule:

$\frac{du}{dx} = \frac{d}{dx}(e^{2x} + e^{-2x})$

$\frac{du}{dx} = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(e^{-2x})$

$\frac{du}{dx} = e^{2x} \cdot \frac{d}{dx}(2x) + e^{-2x} \cdot \frac{d}{dx}(-2x)$

$\frac{du}{dx} = e^{2x} \cdot 2 + e^{-2x} \cdot (-2)$

$\frac{du}{dx} = 2e^{2x} - 2e^{-2x}$

$\frac{du}{dx} = 2(e^{2x} - e^{-2x})$

This gives us the differential $du = 2(e^{2x} - e^{-2x}) \;dx$.

From this, we can express $(e^{2x} - e^{-2x}) \;dx$ as $\frac{1}{2} du$.

Now substitute $u = e^{2x} + e^{-2x}$ and $(e^{2x} - e^{-2x}) \;dx = \frac{1}{2} du$ into the integral:

$I = \int \frac{1}{e^{2x} + e^{-2x}} \cdot (e^{2x} - e^{-2x}) \;dx$

$I = \int \frac{1}{u} \cdot \left(\frac{1}{2} du\right)$

$I = \frac{1}{2} \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{2} (\ln |u| + C') = \frac{1}{2} \ln |u| + \frac{C'}{2}$

Let $C = \frac{C'}{2}$ be the new constant of integration.

$I = \frac{1}{2} \ln |u| + C$

Finally, substitute back $u = e^{2x} + e^{-2x}$ to express the result in terms of $x$:

$I = \frac{1}{2} \ln |e^{2x} + e^{-2x}| + C$

Since $e^{2x} > 0$ and $e^{-2x} > 0$ for all real values of $x$, their sum $e^{2x} + e^{-2x}$ is always positive. Thus, $|e^{2x} + e^{-2x}| = e^{2x} + e^{-2x}$.

$I = \frac{1}{2} \ln (e^{2x} + e^{-2x}) + C$

The integrand $\frac{e^{2x} − e^{−2x}}{e^{2x} + e^{−2x}}$ is the hyperbolic tangent function $\tanh(2x)$.

$\frac{e^{2x} − e^{−2x}}{e^{2x} + e^{−2x}} = \frac{\frac{e^{2x} − e^{−2x}}{2}}{\frac{e^{2x} + e^{−2x}}{2}} = \frac{\sinh(2x)}{\cosh(2x)} = \tanh(2x)$.

So, the integral is $\int \tanh(2x) \;dx$.

We know that $\int \tanh(ax) \;dx = \frac{1}{a} \ln |\cosh(ax)| + C$.

For $a=2$: $\int \tanh(2x) \;dx = \frac{1}{2} \ln |\cosh(2x)| + C$.

Since $\cosh(2x) = \frac{e^{2x} + e^{-2x}}{2}$, and $\frac{e^{2x} + e^{-2x}}{2}$ is always positive, we have $\cosh(2x) > 0$.

$\frac{1}{2} \ln |\cosh(2x)| + C = \frac{1}{2} \ln (\cosh(2x)) + C = \frac{1}{2} \ln \left(\frac{e^{2x} + e^{-2x}}{2}\right) + C$.

Using logarithm properties: $\frac{1}{2} \ln \left(\frac{e^{2x} + e^{-2x}}{2}\right) + C = \frac{1}{2} (\ln (e^{2x} + e^{-2x}) - \ln 2) + C = \frac{1}{2} \ln (e^{2x} + e^{-2x}) - \frac{1}{2}\ln 2 + C$.

The term $-\frac{1}{2}\ln 2$ is a constant and can be absorbed into the integration constant $C$. This shows that both forms of the answer are equivalent.

Therefore, the integral of $\frac{e^{2x} − e^{−2x}}{e^{2x} + e^{−2x}}$ is $\mathbf{\frac{1}{2} \ln (e^{2x} + e^{-2x}) + C}$.

We need to find the integral of the function $\tan^2 (2x - 3)$ with respect to $x$.

Let the integral be $I = \int \tan^2 (2x - 3) \;dx$.

We use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$. Let $\theta = 2x - 3$.

The integrand becomes $\sec^2 (2x - 3) - 1$.

$I = \int (\sec^2 (2x - 3) - 1) \;dx$.

Using the linearity property of integration, we integrate each term separately:

$I = \int \sec^2 (2x - 3) \;dx - \int 1 \;dx$.

First, evaluate $\int \sec^2 (2x - 3) \;dx$ using substitution.

Let $u = 2x - 3$.

Differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(2x - 3) = 2$.

This gives us the differential $du = 2 \;dx$, so $dx = \frac{1}{2} du$.

Substitute into the integral:

$\int \sec^2 (u) \cdot \left(\frac{1}{2} du\right) = \frac{1}{2} \int \sec^2 u \;du$.

The integral of $\sec^2 u$ with respect to $u$ is a standard integral: $\int \sec^2 u \;du = \tan u + C_1$.

So, $\frac{1}{2} \int \sec^2 u \;du = \frac{1}{2} (\tan u + C_1) = \frac{1}{2} \tan u + \frac{C_1}{2}$.

Substitute back $u = 2x - 3$: $\frac{1}{2} \tan (2x - 3) + \frac{C_1}{2}$.

Next, evaluate $\int 1 \;dx$. This is a standard integral:

$\int 1 \;dx = x + C_2$.

Combine the results from both parts of the integral $I = \int \sec^2 (2x - 3) \;dx - \int 1 \;dx$:

$I = \left(\frac{1}{2} \tan (2x - 3) + \frac{C_1}{2}\right) - (x + C_2)$.

$I = \frac{1}{2} \tan (2x - 3) - x + \left(\frac{C_1}{2} - C_2\right)$.

Let $C = \frac{C_1}{2} - C_2$ be the constant of integration.

$I = \frac{1}{2} \tan (2x - 3) - x + C$.

Therefore, the integral of $\tan^2 (2x - 3)$ is $\mathbf{\frac{1}{2} \tan (2x - 3) - x + C}$.

We need to find the integral of the function $\sec^2 (7 - 4x)$ with respect to $x$.

Let the integral be $I = \int \sec^2 (7 - 4x) \;dx$.

We can solve this integral using the method of substitution.

Let $u = 7 - 4x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(7 - 4x) = 0 - 4 = -4$

This gives us the differential $du = -4 \;dx$.

From this, we can express $dx$ as $-\frac{1}{4} du$.

Now substitute $u = 7 - 4x$ and $dx = -\frac{1}{4} du$ into the integral:

$I = \int \sec^2 (u) \cdot \left(-\frac{1}{4} du\right)$

$I = -\frac{1}{4} \int \sec^2 u \;du$

The integral of $\sec^2 u$ with respect to $u$ is a standard integral:

$\int \sec^2 u \;du = \tan u + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = -\frac{1}{4} (\tan u + C') = -\frac{1}{4} \tan u - \frac{C'}{4}$

Let $C = -\frac{C'}{4}$ be the new constant of integration.

$I = -\frac{1}{4} \tan u + C$

Finally, substitute back $u = 7 - 4x$ to express the result in terms of $x$:

$I = -\frac{1}{4} \tan (7 - 4x) + C$

Therefore, the integral of $\sec^2 (7 - 4x)$ is $\mathbf{-\frac{1}{4} \tan (7 - 4x) + C}$.

We need to find the integral of the function $\frac{\sin^{-1} x}{\sqrt{1 - x^2}}$ with respect to $x$.

Let the integral be $I = \int \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \;dx$.

We can rewrite the integrand as $\sin^{-1} x \cdot \frac{1}{\sqrt{1 - x^2}}$.

$I = \int \sin^{-1} x \cdot \frac{1}{\sqrt{1 - x^2}} \;dx$.

We can solve this integral using the method of substitution.

Let $u = \sin^{-1} x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$

This gives us the differential $du = \frac{1}{\sqrt{1 - x^2}} \;dx$.

Now substitute $u = \sin^{-1} x$ and $du = \frac{1}{\sqrt{1 - x^2}} \;dx$ into the integral:

$I = \int u \;du$

The integral of $u$ with respect to $u$ is a standard integral using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u \;du = \frac{u^{1+1}}{1+1} + C' = \frac{u^2}{2} + C'$

where $C'$ is the constant of integration.

Let $C = C'$ be the new constant of integration.

$I = \frac{u^2}{2} + C$

Finally, substitute back $u = \sin^{-1} x$ to express the result in terms of $x$:

$I = \frac{(\sin^{-1} x)^2}{2} + C$

Therefore, the integral of $\frac{\sin^{-1} x}{\sqrt{1 - x^2}}$ is $\mathbf{\frac{(\sin^{-1} x)^2}{2} + C}$.

We need to find the integral of the function $\frac{2\cos x − 3 \sin x}{6\cos x + 4\sin x}$ with respect to $x$.

Let the integral be $I = \int \frac{2\cos x − 3 \sin x}{6\cos x + 4\sin x} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 6\cos x + 4\sin x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(6\cos x + 4\sin x)$

$\frac{du}{dx} = 6 \frac{d}{dx}(\cos x) + 4 \frac{d}{dx}(\sin x)$

$\frac{du}{dx} = 6 (-\sin x) + 4 (\cos x)$

$\frac{du}{dx} = -6 \sin x + 4 \cos x$

$\frac{du}{dx} = 4 \cos x - 6 \sin x$

The numerator of the integrand is $2\cos x - 3 \sin x$. Notice that the derivative of the denominator is a multiple of the numerator.

We can write $4 \cos x - 6 \sin x = 2(2 \cos x - 3 \sin x)$.

This gives us the differential $du = (4 \cos x - 6 \sin x) \;dx = 2(2 \cos x - 3 \sin x) \;dx$.

From this, we can express $(2 \cos x - 3 \sin x) \;dx$ as $\frac{1}{2} du$.

Now substitute $u = 6\cos x + 4\sin x$ and $(2 \cos x - 3 \sin x) \;dx = \frac{1}{2} du$ into the integral:

$I = \int \frac{1}{6\cos x + 4\sin x} \cdot (2\cos x − 3 \sin x) \;dx$

$I = \int \frac{1}{u} \cdot \left(\frac{1}{2} du\right)$

$I = \frac{1}{2} \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{2} (\ln |u| + C') = \frac{1}{2} \ln |u| + \frac{C'}{2}$

Let $C = \frac{C'}{2}$ be the new constant of integration.

$I = \frac{1}{2} \ln |u| + C$

Finally, substitute back $u = 6\cos x + 4\sin x$ to express the result in terms of $x$:

$I = \frac{1}{2} \ln |6\cos x + 4\sin x| + C$

Therefore, the integral of $\frac{2\cos x − 3 \sin x}{6\cos x + 4\sin x}$ is $\mathbf{\frac{1}{2} \ln |6\cos x + 4\sin x| + C}$.

We need to find the integral of the function $\frac{1}{\cos^2 x (1 − \tan x)^2}$ with respect to $x$.

Let the integral be $I = \int \frac{1}{\cos^2 x (1 − \tan x)^2} \;dx$.

We can rewrite $\frac{1}{\cos^2 x}$ as $\sec^2 x$.

The integrand becomes $\frac{\sec^2 x}{(1 − \tan x)^2}$.

$I = \int \frac{\sec^2 x}{(1 − \tan x)^2} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 - \tan x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 - \tan x) = 0 - \sec^2 x = -\sec^2 x$

This gives us the differential $du = -\sec^2 x \;dx$.

From this, we can express $\sec^2 x \;dx$ as $-du$.

Now substitute $u = 1 - \tan x$ and $\sec^2 x \;dx = -du$ into the integral:

$I = \int \frac{1}{(1 − \tan x)^2} \cdot (\sec^2 x \;dx)$

$I = \int \frac{1}{u^2} \cdot (-du)$

$I = - \int \frac{1}{u^2} \;du$

Rewrite $\frac{1}{u^2}$ using negative exponents: $\frac{1}{u^2} = u^{-2}$.

$I = - \int u^{-2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$). Here $n = -2$:

$\int u^{-2} \;du = \frac{u^{-2+1}}{-2+1} + C' = \frac{u^{-1}}{-1} + C' = -u^{-1} + C' = -\frac{1}{u} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = - \left(-\frac{1}{u} + C'\right) = \frac{1}{u} - C'$

Let $C = -C'$ be the new constant of integration.

$I = \frac{1}{u} + C$

Finally, substitute back $u = 1 - \tan x$ to express the result in terms of $x$:

$I = \frac{1}{1 - \tan x} + C$

Therefore, the integral of $\frac{1}{\cos^2 x (1 − \tan x)^2}$ is $\mathbf{\frac{1}{1 - \tan x} + C}$.

We need to find the integral of the function $\frac{\cos \sqrt{x}}{\sqrt{x}}$ with respect to $x$.

Let the integral be $I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} \;dx$.

We can rewrite the integrand as $\cos \sqrt{x} \cdot \frac{1}{\sqrt{x}}$.

$I = \int \cos \sqrt{x} \cdot \frac{1}{\sqrt{x}} \;dx$.

We can solve this integral using the method of substitution.

Let $u = \sqrt{x}$.

Differentiate $u$ with respect to $x$. Rewrite $\sqrt{x}$ as $x^{1/2}$:

$u = x^{1/2}$

$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

This gives us the differential $du = \frac{1}{2\sqrt{x}} \;dx$.

From this, we can express $\frac{1}{\sqrt{x}} \;dx$ as $2 \;du$.

Now substitute $u = \sqrt{x}$ and $\frac{1}{\sqrt{x}} \;dx = 2 \;du$ into the integral:

$I = \int \cos u \cdot (2 \;du)$

$I = 2 \int \cos u \;du$

The integral of $\cos u$ with respect to $u$ is a standard integral:

$\int \cos u \;du = \sin u + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = 2 (\sin u + C') = 2 \sin u + 2C'$

Let $C = 2C'$ be the new constant of integration.

$I = 2 \sin u + C$

Finally, substitute back $u = \sqrt{x}$ to express the result in terms of $x$:

$I = 2 \sin \sqrt{x} + C$

Therefore, the integral of $\frac{\cos \sqrt{x}}{\sqrt{x}}$ is $\mathbf{2 \sin \sqrt{x} + C}$.

We need to find the integral of the function $\sqrt{\sin 2x} \cos 2x$ with respect to $x$.

Let the integral be $I = \int \sqrt{\sin 2x} \cos 2x \;dx$.

We can rewrite the square root as a power: $\sqrt{\sin 2x} = (\sin 2x)^{1/2}$.

$I = \int (\sin 2x)^{1/2} \cos 2x \;dx$.

We can solve this integral using the method of substitution.

Let $u = \sin 2x$.

Differentiate $u$ with respect to $x$ using the chain rule:

$\frac{du}{dx} = \frac{d}{dx}(\sin 2x) = \cos 2x \cdot \frac{d}{dx}(2x)$

$\frac{du}{dx} = \cos 2x \cdot 2 = 2 \cos 2x$

This gives us the differential $du = 2 \cos 2x \;dx$.

From this, we can express $\cos 2x \;dx$ as $\frac{1}{2} du$.

Now substitute $u = \sin 2x$ and $\cos 2x \;dx = \frac{1}{2} du$ into the integral:

$I = \int u^{1/2} \cdot \left(\frac{1}{2} du\right)$

$I = \frac{1}{2} \int u^{1/2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$). Here $n = 1/2$:

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C' = \frac{u^{3/2}}{3/2} + C' = \frac{2}{3}u^{3/2} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{2} \left(\frac{2}{3}u^{3/2} + C'\right) = \frac{2}{6}u^{3/2} + \frac{C'}{2} = \frac{1}{3}u^{3/2} + \frac{C'}{2}$

Let $C = \frac{C'}{2}$ be the new constant of integration.

$I = \frac{1}{3}u^{3/2} + C$

Finally, substitute back $u = \sin 2x$ to express the result in terms of $x$:

$I = \frac{1}{3}(\sin 2x)^{3/2} + C = \frac{1}{3}\sqrt{(\sin 2x)^3} + C = \frac{1}{3}(\sin 2x)\sqrt{\sin 2x} + C$

Therefore, the integral of $\sqrt{\sin 2x} \cos 2x$ is $\mathbf{\frac{1}{3}(\sin 2x)^{\frac{3}{2}} + C}$.

We need to find the integral of the function $\frac{\cos x}{\sqrt{1 + \sin x}}$ with respect to $x$.

Let the integral be $I = \int \frac{\cos x}{\sqrt{1 + \sin x}} \;dx$.

We can rewrite the integrand as $\frac{1}{\sqrt{1 + \sin x}} \cdot \cos x$.

$I = \int \frac{1}{\sqrt{1 + \sin x}} \cdot \cos x \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 + \sin x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + \sin x) = 0 + \cos x = \cos x$

This gives us the differential $du = \cos x \;dx$.

Now substitute $u = 1 + \sin x$ and $du = \cos x \;dx$ into the integral:

$I = \int \frac{1}{\sqrt{u}} \;du$

Rewrite $\frac{1}{\sqrt{u}}$ using negative exponents: $\frac{1}{\sqrt{u}} = u^{-1/2}$.

$I = \int u^{-1/2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$). Here $n = -1/2$:

$\int u^{-1/2} \;du = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C' = \frac{u^{1/2}}{1/2} + C' = 2u^{1/2} + C'$

where $C'$ is the constant of integration.

Let $C = C'$ be the new constant of integration.

$I = 2u^{1/2} + C$

Finally, substitute back $u = 1 + \sin x$ to express the result in terms of $x$:

$I = 2(1 + \sin x)^{1/2} + C = 2\sqrt{1 + \sin x} + C$

Therefore, the integral of $\frac{\cos x}{\sqrt{1 + \sin x}}$ is $\mathbf{2\sqrt{1 + \sin x} + C}$.

We need to find the integral of the function $\cot x \log \sin x$ with respect to $x$.

Let the integral be $I = \int \cot x \log \sin x \;dx$.

We can rewrite the integrand as $(\log \sin x) \cdot \cot x$.

$I = \int (\log \sin x) \cdot \cot x \;dx$.

We can solve this integral using the method of substitution.

Let $u = \log \sin x$.

Differentiate $u$ with respect to $x$ using the chain rule:

$\frac{du}{dx} = \frac{d}{dx}(\log \sin x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$

$\frac{du}{dx} = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x$

This gives us the differential $du = \cot x \;dx$.

Now substitute $u = \log \sin x$ and $du = \cot x \;dx$ into the integral:

$I = \int u \;du$

The integral of $u$ with respect to $u$ is a standard integral using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u \;du = \frac{u^{1+1}}{1+1} + C' = \frac{u^2}{2} + C'$

where $C'$ is the constant of integration.

Let $C = C'$ be the new constant of integration.

$I = \frac{u^2}{2} + C$

Finally, substitute back $u = \log \sin x$ to express the result in terms of $x$:

$I = \frac{(\log \sin x)^2}{2} + C$

Therefore, the integral of $\cot x \log \sin x$ is $\mathbf{\frac{(\log \sin x)^2}{2} + C}$.

We need to find the integral of the function $\frac{\sin x}{1 + \cos x}$ with respect to $x$.

Let the integral be $I = \int \frac{\sin x}{1 + \cos x} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 + \cos x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + \cos x) = 0 - \sin x = -\sin x$

This gives us the differential $du = -\sin x \;dx$.

From this, we can express $\sin x \;dx$ as $-du$.

Now substitute $u = 1 + \cos x$ and $\sin x \;dx = -du$ into the integral:

$I = \int \frac{1}{1 + \cos x} \cdot (\sin x \;dx)$

$I = \int \frac{1}{u} \cdot (-du)$

$I = - \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = - (\ln |u| + C') = -\ln |u| - C'$

Let $C = -C'$ be the new constant of integration.

$I = -\ln |u| + C$

Finally, substitute back $u = 1 + \cos x$ to express the result in terms of $x$:

$I = -\ln |1 + \cos x| + C$

Therefore, the integral of $\frac{\sin x}{1 + \cos x}$ is $\mathbf{-\ln |1 + \cos x| + C}$.

We need to find the integral of the function $\frac{\sin x}{(1 + \cos x)^2}$ with respect to $x$.

Let the integral be $I = \int \frac{\sin x}{(1 + \cos x)^2} \;dx$.

We can rewrite the integrand as $\frac{1}{(1 + \cos x)^2} \cdot \sin x$.

$I = \int \frac{1}{(1 + \cos x)^2} \cdot \sin x \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 + \cos x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + \cos x) = 0 - \sin x = -\sin x$

This gives us the differential $du = -\sin x \;dx$.

From this, we can express $\sin x \;dx$ as $-du$.

Now substitute $u = 1 + \cos x$ and $\sin x \;dx = -du$ into the integral:

$I = \int \frac{1}{u^2} \cdot (-du)$

$I = - \int \frac{1}{u^2} \;du$

Rewrite $\frac{1}{u^2}$ using negative exponents: $\frac{1}{u^2} = u^{-2}$.

$I = - \int u^{-2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$). Here $n = -2$:

$\int u^{-2} \;du = \frac{u^{-2+1}}{-2+1} + C' = \frac{u^{-1}}{-1} + C' = -u^{-1} + C' = -\frac{1}{u} + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = - \left(-\frac{1}{u} + C'\right) = \frac{1}{u} - C'$

Let $C = -C'$ be the new constant of integration.

$I = \frac{1}{u} + C$

Finally, substitute back $u = 1 + \cos x$ to express the result in terms of $x$:

$I = \frac{1}{1 + \cos x} + C$

Therefore, the integral of $\frac{\sin x}{(1 + \cos x)^2}$ is $\mathbf{\frac{1}{1 + \cos x} + C}$.

We need to find the integral of the function $\frac{1}{1 + \cot x}$ with respect to $x$.

Let the integral be $I = \int \frac{1}{1 + \cot x} \;dx$.

Rewrite $\cot x$ as $\frac{\cos x}{\sin x}$:

$\frac{1}{1 + \frac{\cos x}{\sin x}} = \frac{1}{\frac{\sin x + \cos x}{\sin x}} = \frac{\sin x}{\sin x + \cos x}$.

The integral is $I = \int \frac{\sin x}{\sin x + \cos x} \;dx$.

This is a common type of integral. We can use a trick: write the numerator in terms of the denominator and its derivative.

Let the numerator be $A(\sin x + \cos x) + B \frac{d}{dx}(\sin x + \cos x)$.

$\frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$.

So, $\sin x = A(\sin x + \cos x) + B (\cos x - \sin x)$

$\sin x = A \sin x + A \cos x + B \cos x - B \sin x$

$\sin x = (A - B) \sin x + (A + B) \cos x$.

Comparing coefficients of $\sin x$ and $\cos x$ on both sides:

Adding (1) and (2): $2A = 1 \implies A = \frac{1}{2}$.

Substituting $A = \frac{1}{2}$ into (2): $\frac{1}{2} + B = 0 \implies B = -\frac{1}{2}$.

So, $\sin x = \frac{1}{2}(\sin x + \cos x) - \frac{1}{2}(\cos x - \sin x)$.

The integrand becomes $\frac{\frac{1}{2}(\sin x + \cos x) - \frac{1}{2}(\cos x - \sin x)}{\sin x + \cos x} = \frac{1}{2} \frac{\sin x + \cos x}{\sin x + \cos x} - \frac{1}{2} \frac{\cos x - \sin x}{\sin x + \cos x}$.

For $\sin x + \cos x \neq 0$, this simplifies to $\frac{1}{2} - \frac{1}{2} \frac{\cos x - \sin x}{\sin x + \cos x}$.

Now, integrate this expression:

$I = \int \left(\frac{1}{2} - \frac{1}{2} \frac{\cos x - \sin x}{\sin x + \cos x}\right) \;dx = \int \frac{1}{2} \;dx - \int \frac{1}{2} \frac{\cos x - \sin x}{\sin x + \cos x} \;dx$

$\int \frac{1}{2} \;dx = \frac{1}{2}x + C_1$.

For the second integral, let $v = \sin x + \cos x$. Then $\frac{dv}{dx} = \cos x - \sin x$, so $dv = (\cos x - \sin x) \;dx$.

The second integral becomes $- \int \frac{1}{2} \frac{dv}{v} = -\frac{1}{2} \int \frac{1}{v} \;dv$.

$-\frac{1}{2} \int \frac{1}{v} \;dv = -\frac{1}{2} \ln |v| + C_2$.

Substitute back $v = \sin x + \cos x$:

$-\frac{1}{2} \ln |\sin x + \cos x| + C_2$.

Combining the results:

$I = \frac{1}{2}x - \frac{1}{2} \ln |\sin x + \cos x| + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral of $\frac{1}{1 + \cot x}$ is $\mathbf{\frac{1}{2}x - \frac{1}{2} \ln |\sin x + \cos x| + C}$.

We need to find the integral of the function $\frac{1}{1 − \tan x}$ with respect to $x$.

Let the integral be $I = \int \frac{1}{1 − \tan x} \;dx$.

Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$:

$\frac{1}{1 - \frac{\sin x}{\cos x}} = \frac{1}{\frac{\cos x - \sin x}{\cos x}} = \frac{\cos x}{\cos x - \sin x}$.

The integral is $I = \int \frac{\cos x}{\cos x - \sin x} \;dx$.

This is a common type of integral. We can use a trick: write the numerator in terms of the denominator and its derivative.

Let the numerator be $A(\cos x - \sin x) + B \frac{d}{dx}(\cos x - \sin x)$.

$\frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x = -(\sin x + \cos x)$.

So, $\cos x = A(\cos x - \sin x) + B (-(\sin x + \cos x))$

$\cos x = A \cos x - A \sin x - B \sin x - B \cos x$

$\cos x = (A - B) \cos x + (-A - B) \sin x$.

Comparing coefficients of $\cos x$ and $\sin x$ on both sides:

Adding (1) and (2): $(A - B) + (-A - B) = 1 + 0 \implies -2B = 1 \implies B = -\frac{1}{2}$.

Substituting $B = -\frac{1}{2}$ into (1): $A - (-\frac{1}{2}) = 1 \implies A + \frac{1}{2} = 1 \implies A = 1 - \frac{1}{2} = \frac{1}{2}$.

So, $\cos x = \frac{1}{2}(\cos x - \sin x) - \frac{1}{2}(\sin x + \cos x)$.

The integrand becomes $\frac{\frac{1}{2}(\cos x - \sin x) - \frac{1}{2}(\sin x + \cos x)}{\cos x - \sin x} = \frac{1}{2} \frac{\cos x - \sin x}{\cos x - \sin x} - \frac{1}{2} \frac{\sin x + \cos x}{\cos x - \sin x}$.

For $\cos x - \sin x \neq 0$, this simplifies to $\frac{1}{2} - \frac{1}{2} \frac{\sin x + \cos x}{\cos x - \sin x}$.

Now, integrate this expression:

$I = \int \left(\frac{1}{2} - \frac{1}{2} \frac{\sin x + \cos x}{\cos x - \sin x}\right) \;dx = \int \frac{1}{2} \;dx - \int \frac{1}{2} \frac{\sin x + \cos x}{\cos x - \sin x} \;dx$

$\int \frac{1}{2} \;dx = \frac{1}{2}x + C_1$.

For the second integral, let $v = \cos x - \sin x$. Then $\frac{dv}{dx} = -\sin x - \cos x = -(\sin x + \cos x)$, so $dv = -(\sin x + \cos x) \;dx$, which means $(\sin x + \cos x) \;dx = -dv$.

The second integral becomes $- \int \frac{1}{2} \frac{-dv}{v} = \frac{1}{2} \int \frac{1}{v} \;dv$.

$\frac{1}{2} \int \frac{1}{v} \;dv = \frac{1}{2} \ln |v| + C_2$.

Substitute back $v = \cos x - \sin x$:

$\frac{1}{2} \ln |\cos x - \sin x| + C_2$.

Combining the results:

$I = \frac{1}{2}x + \frac{1}{2} \ln |\cos x - \sin x| + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral of $\frac{1}{1 − \tan x}$ is $\mathbf{\frac{1}{2}x + \frac{1}{2} \ln |\cos x - \sin x| + C}$.

We need to find the integral of the function $\frac{\sqrt{\tan x}}{\sin x \cos x}$ with respect to $x$.

Let the integral be $I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \;dx$.

We can simplify the integrand by manipulating the denominator. Divide the denominator by $\cos x$ and multiply the numerator by $\cos x$ (or multiply the expression by $\frac{\sec x}{\sec x}$):

$\frac{\sqrt{\tan x}}{\sin x \cos x} = \frac{\sqrt{\tan x}}{\sin x \cos x} \cdot \frac{\frac{1}{\cos x}}{\frac{1}{\cos x}} = \frac{\sqrt{\tan x} \frac{1}{\cos x}}{\frac{\sin x \cos x}{\cos x}} \cdot \frac{1}{\cos x} = \frac{\sqrt{\tan x} \sec x}{\sin x} \sec x$ This does not seem right.

Let's divide both the numerator and the denominator by $\cos^2 x$:

$\frac{\frac{\sqrt{\tan x}}{\cos^2 x}}{\frac{\sin x \cos x}{\cos^2 x}} = \frac{\sqrt{\tan x} \sec^2 x}{\tan x}$

Rewrite $\sqrt{\tan x}$ as $(\tan x)^{1/2}$ and $\tan x$ as $(\tan x)^1$:

$\frac{(\tan x)^{1/2} \sec^2 x}{(\tan x)^1} = (\tan x)^{1/2 - 1} \sec^2 x = (\tan x)^{-1/2} \sec^2 x = \frac{\sec^2 x}{\sqrt{\tan x}}$.

The integral is $I = \int \frac{\sec^2 x}{\sqrt{\tan x}} \;dx$.

We can solve this integral using the method of substitution.

Let $u = \tan x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

This gives us the differential $du = \sec^2 x \;dx$.

Now substitute $u = \tan x$ and $du = \sec^2 x \;dx$ into the integral:

$I = \int \frac{1}{\sqrt{\tan x}} \cdot (\sec^2 x \;dx)$

$I = \int \frac{1}{\sqrt{u}} \;du$

Rewrite $\frac{1}{\sqrt{u}}$ using negative exponents: $\frac{1}{\sqrt{u}} = u^{-1/2}$.

$I = \int u^{-1/2} \;du$

Apply the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$ (for $n \neq -1$). Here $n = -1/2$:

$\int u^{-1/2} \;du = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C' = \frac{u^{1/2}}{1/2} + C' = 2u^{1/2} + C'$

where $C'$ is the constant of integration.

Let $C = C'$ be the new constant of integration.

$I = 2u^{1/2} + C$

Finally, substitute back $u = \tan x$ to express the result in terms of $x$:

$I = 2(\tan x)^{1/2} + C = 2\sqrt{\tan x} + C$

Therefore, the integral of $\frac{\sqrt{\tan x}}{\sin x \cos x}$ is $\mathbf{2\sqrt{\tan x} + C}$.

We need to find the integral of the function $\frac{(1 + \log x)^2}{x}$ with respect to $x$.

Let the integral be $I = \int \frac{(1 + \log x)^2}{x} \;dx$.

We can rewrite the integrand as $(1 + \log x)^2 \cdot \frac{1}{x}$.

$I = \int (1 + \log x)^2 \cdot \frac{1}{x} \;dx$.

We can solve this integral using the method of substitution.

Let $u = 1 + \log x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + \log x) = 0 + \frac{1}{x} = \frac{1}{x}$

This gives us the differential $du = \frac{1}{x} \;dx$.

Now substitute $u = 1 + \log x$ and $du = \frac{1}{x} \;dx$ into the integral:

$I = \int u^2 \;du$

The integral of $u^2$ with respect to $u$ is a standard integral using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C'$:

$\int u^2 \;du = \frac{u^{2+1}}{2+1} + C' = \frac{u^3}{3} + C'$

where $C'$ is the constant of integration.

Let $C = C'$ be the new constant of integration.

$I = \frac{u^3}{3} + C$

Finally, substitute back $u = 1 + \log x$ to express the result in terms of $x$:

$I = \frac{(1 + \log x)^3}{3} + C$

Therefore, the integral of $\frac{(1 + \log x)^2}{x}$ is $\mathbf{\frac{(1 + \log x)^3}{3} + C}$.

We need to find the integral of the function $\frac{(x + 1) (x + \log x)^2}{x}$ with respect to $x$.

Let the integral be $I = \int \frac{(x + 1) (x + \log x)^2}{x} \;dx$.

We can rewrite the integrand by splitting the fraction:

$\frac{(x + 1)}{x} (x + \log x)^2 = \left(\frac{x}{x} + \frac{1}{x}\right) (x + \log x)^2 = \left(1 + \frac{1}{x}\right) (x + \log x)^2$

So, the integral is $I = \int \left(1 + \frac{1}{x}\right) (x + \log x)^2 \;dx$.

We can solve this integral using the method of substitution.

Let $u = x + \log x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x + \log x) = \frac{d}{dx}(x) + \frac{d}{dx}(\log x) = 1 + \frac{1}{x}$

This gives us the differential $du = \left(1 + \frac{1}{x}\right) \;dx$.

Now substitute $u = x + \log x$ and $du = \left(1 + \frac{1}{x}\right) \;dx$ into the integral:

$I = \int (x + \log x)^2 \cdot \left(1 + \frac{1}{x}\right) \;dx$

$I = \int u^2 \;du$

The integral of $u^2$ with respect to $u$ is a standard integral using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C$:

$\int u^2 \;du = \frac{u^{2+1}}{2+1} + C = \frac{u^3}{3} + C$

where $C$ is the constant of integration.

Substitute back $u = x + \log x$ to express the result in terms of $x$:

$I = \frac{(x + \log x)^3}{3} + C$

Therefore, the integral of $\frac{(x + 1) (x + \log x)^2}{x}$ is $\mathbf{\frac{(x + \log x)^3}{3} + C}$.

We need to find the integral of the function $\frac{x^3 \sin (\tan^{-1} x^4)}{1 + x^8}$ with respect to $x$.

Let the integral be $I = \int \frac{x^3 \sin (\tan^{-1} x^4)}{1 + x^8} \;dx$.

We can rewrite the integrand to group terms for substitution:

$\frac{x^3 \sin (\tan^{-1} x^4)}{1 + x^8} = \sin (\tan^{-1} x^4) \cdot \frac{x^3}{1 + (x^4)^2}$

We can solve this integral using the method of substitution.

Consider the term $\tan^{-1} x^4$. Its derivative is related to the remaining part of the integrand.

Let $u = \tan^{-1} x^4$.

Differentiate $u$ with respect to $x$ using the chain rule. Recall $\frac{d}{dx}(\tan^{-1} v) = \frac{1}{1 + v^2} \frac{dv}{dx}$. Here $v = x^4$.

$\frac{du}{dx} = \frac{d}{dx}(\tan^{-1} x^4) = \frac{1}{1 + (x^4)^2} \cdot \frac{d}{dx}(x^4)$

$\frac{du}{dx} = \frac{1}{1 + x^8} \cdot 4x^{4-1}$

$\frac{du}{dx} = \frac{4x^3}{1 + x^8}$

This gives us the differential $du = \frac{4x^3}{1 + x^8} \;dx$.

From this, we can express $\frac{x^3}{1 + x^8} \;dx$ as $\frac{1}{4} du$.

Now substitute $u = \tan^{-1} x^4$ and $\frac{x^3}{1 + x^8} \;dx = \frac{1}{4} du$ into the integral:

$I = \int \sin (\tan^{-1} x^4) \cdot \left(\frac{x^3}{1 + x^8} \;dx\right)$

$I = \int \sin u \cdot \left(\frac{1}{4} du\right)$

$I = \frac{1}{4} \int \sin u \;du$

The integral of $\sin u$ with respect to $u$ is a standard integral:

$\int \sin u \;du = -\cos u + C'$

where $C'$ is the constant of integration.

Substitute this result back into the expression for $I$:

$I = \frac{1}{4} (-\cos u + C') = -\frac{1}{4}\cos u + \frac{C'}{4}$

Let $C = \frac{C'}{4}$ be the new constant of integration.

$I = -\frac{1}{4}\cos u + C$

Finally, substitute back $u = \tan^{-1} x^4$ to express the result in terms of $x$:

$I = -\frac{1}{4}\cos (\tan^{-1} x^4) + C$

Therefore, the integral of $\frac{x^3 \sin (\tan^{-1} x^4)}{1 + x^8}$ is $\mathbf{-\frac{1}{4}\cos (\tan^{-1} x^4) + C}$.

We need to find the integral of the function $\frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x}$ with respect to $x$.

Let the integral be $I = \int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x} \;dx$.

We can solve this integral using the method of substitution.

Let $u = x^{10} + 10^x$.

Differentiate $u$ with respect to $x$. Recall that $\frac{d}{dx}(x^n) = nx^{n-1}$ and $\frac{d}{dx}(a^x) = a^x \ln a$ (where $\ln a = \log_e a$).

$\frac{du}{dx} = \frac{d}{dx}(x^{10} + 10^x)$

$\frac{du}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(10^x)$

$\frac{du}{dx} = 10x^{10-1} + 10^x \log_e 10$

$\frac{du}{dx} = 10x^9 + 10^x \log_e 10$

This gives us the differential $du = (10x^9 + 10^x \log_e 10) \;dx$.

Now observe that the numerator of the integrand is exactly this differential $du$. Substitute $u = x^{10} + 10^x$ and $du = (10x^9 + 10^x \log_e 10) \;dx$ into the integral:

$I = \int \frac{du}{u}$

$I = \int \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is a standard integral:

$\int \frac{1}{u} \;du = \ln |u| + C$

where $C$ is the constant of integration. Note that $\log$ typically refers to $\ln$ in calculus context, or specifically $\log_e$.

Substitute back $u = x^{10} + 10^x$ to express the result in terms of $x$:

$I = \ln |x^{10} + 10^x| + C$

Since $x^{10} + 10^x$ is positive for most real values of $x$, the absolute value is often omitted in multiple choice options.

$I = \ln (x^{10} + 10^x) + C$

Comparing this result with the given options, and interpreting "log" as the natural logarithm, we find that it matches option (D).

The correct answer is (D) $\mathbf{\log (10^x + x^{10}) + C}$.

We need to find the integral of the function $\frac{1}{\sin^2 x \cos^2 x}$ with respect to $x$.

Let the integral be $I = \int \frac{1}{\sin^2 x \cos^2 x} \;dx$.

We can simplify the integrand using trigonometric identities.

Recall the identity $\sin^2 x + \cos^2 x = 1$. We can replace the numerator '1' with this identity:

$I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \;dx$

Now, split the fraction into two terms:

$I = \int \left(\frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x}\right) \;dx$

Simplify each term:

$\frac{\sin^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$

$\frac{\cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x} = \text{cosec}^2 x$

So, the integral becomes:

$I = \int (\sec^2 x + \text{cosec}^2 x) \;dx$

Using the linearity property of integration, we integrate each term separately:

$I = \int \sec^2 x \;dx + \int \text{cosec}^2 x \;dx$

Using the standard integral formulas $\int \sec^2 x \;dx = \tan x + C_1$ and $\int \text{cosec}^2 x \;dx = -\cot x + C_2$:

$I = (\tan x + C_1) + (-\cot x + C_2)$

$I = \tan x - \cot x + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

$I = \tan x - \cot x + C$

Comparing this result with the given options, we find that it matches option (B).

Alternatively, we can use the identity $\sin 2x = 2 \sin x \cos x$.

$\sin^2 x \cos^2 x = (\sin x \cos x)^2 = \left(\frac{\sin 2x}{2}\right)^2 = \frac{\sin^2 2x}{4}$.

The integral becomes $I = \int \frac{1}{\frac{\sin^2 2x}{4}} \;dx = \int \frac{4}{\sin^2 2x} \;dx = 4 \int \text{cosec}^2 2x \;dx$.

Let $u = 2x$. Then $du = 2 \;dx$, so $dx = \frac{1}{2} du$.

$I = 4 \int \text{cosec}^2 u \cdot \frac{1}{2} \;du = 2 \int \text{cosec}^2 u \;du$.

$I = 2 (-\cot u) + C' = -2 \cot u + C'$.

Substitute back $u = 2x$:

$I = -2 \cot 2x + C'$.

Using the identity $\cot 2x = \frac{\cot^2 x - 1}{2 \cot x}$ or $\cot 2x = \frac{1 - \tan^2 x}{2 \tan x}$, we have:

$-2 \cot 2x = -2 \left(\frac{\cos 2x}{\sin 2x}\right) = -2 \left(\frac{\cos^2 x - \sin^2 x}{2 \sin x \cos x}\right) = -\left(\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}\right)$

$= -\left(\frac{\cos^2 x}{\sin x \cos x} - \frac{\sin^2 x}{\sin x \cos x}\right) = -\left(\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}\right) = -(\cot x - \tan x) = \tan x - \cot x$.

So, $-2 \cot 2x = \tan x - \cot x$.

Thus, the result from the second method is also $\tan x - \cot x + C$ (by letting $C = C'$).

Both methods yield the same form of the anti derivative. Comparing with the options, the correct answer is (B).

The correct answer is (B) $\mathbf{\tan x – \cot x + C}$.

We will find the integrals using trigonometric identities.

(i) $\int \cos^2 x \;dx$

We use the double angle identity $\cos 2x = 2\cos^2 x - 1$, which gives $\cos^2 x = \frac{1 + \cos 2x}{2}$.

The integral becomes:

$\int \frac{1 + \cos 2x}{2} \;dx = \int \left(\frac{1}{2} + \frac{1}{2}\cos 2x\right) \;dx$

Using the linearity property, integrate term by term:

$\int \frac{1}{2} \;dx + \int \frac{1}{2}\cos 2x \;dx$

$\int \frac{1}{2} \;dx = \frac{1}{2}x + C_1$

For $\int \frac{1}{2}\cos 2x \;dx$, take the constant out: $\frac{1}{2} \int \cos 2x \;dx$.

Use the substitution $u = 2x$, so $du = 2 \;dx$, or $dx = \frac{1}{2} du$.

$\frac{1}{2} \int \cos u \cdot \frac{1}{2} \;du = \frac{1}{4} \int \cos u \;du = \frac{1}{4}\sin u + C_2$.

Substitute back $u = 2x$: $\frac{1}{4}\sin 2x + C_2$.

Combining the results:

$\int \cos^2 x \;dx = \frac{1}{2}x + \frac{1}{4}\sin 2x + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, $\int \cos^2 x \;dx = \mathbf{\frac{x}{2} + \frac{\sin 2x}{4} + C}$.

(ii) $\int \sin 2x \cos 3x \;dx$

We use the product-to-sum trigonometric identity: $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$.

Let $A = 2x$ and $B = 3x$.

$\sin 2x \cos 3x = \frac{1}{2}[\sin(2x+3x) + \sin(2x-3x)] = \frac{1}{2}[\sin 5x + \sin(-x)]$

Since $\sin(-x) = -\sin x$, this becomes $\frac{1}{2}[\sin 5x - \sin x]$.

The integral becomes:

$\int \frac{1}{2}[\sin 5x - \sin x] \;dx = \frac{1}{2} \int (\sin 5x - \sin x) \;dx$

Using the linearity property, integrate term by term:

$\frac{1}{2} \left( \int \sin 5x \;dx - \int \sin x \;dx \right)$

Use the formula $\int \sin (ax) \;dx = -\frac{1}{a}\cos (ax) + C'$.

$\int \sin 5x \;dx = -\frac{1}{5}\cos 5x + C_1$

$\int \sin x \;dx = -\cos x + C_2$

Substituting back:

$\frac{1}{2} \left( \left(-\frac{1}{5}\cos 5x + C_1\right) - (-\cos x + C_2) \right)$

$\frac{1}{2} \left(-\frac{1}{5}\cos 5x + \cos x + C_1 - C_2\right)$

$-\frac{1}{10}\cos 5x + \frac{1}{2}\cos x + \frac{1}{2}(C_1 - C_2)$

Let $C = \frac{1}{2}(C_1 - C_2)$ be the constant of integration.

Therefore, $\int \sin 2x \cos 3x \;dx = \mathbf{-\frac{1}{10}\cos 5x + \frac{1}{2}\cos x + C}$.

(iii) $\int \sin^3 x \;dx$

We can rewrite $\sin^3 x$ as $\sin^2 x \sin x$. Use the identity $\sin^2 x = 1 - \cos^2 x$.

The integrand becomes $(1 - \cos^2 x) \sin x = \sin x - \cos^2 x \sin x$.

The integral is $\int (\sin x - \cos^2 x \sin x) \;dx$.

Using the linearity property, integrate term by term:

$\int \sin x \;dx - \int \cos^2 x \sin x \;dx$

$\int \sin x \;dx = -\cos x + C_1$.

For the second integral, $\int \cos^2 x \sin x \;dx$, use substitution. Let $u = \cos x$. Then $du = -\sin x \;dx$, so $\sin x \;dx = -du$.

$\int u^2 (-du) = - \int u^2 \;du = -\frac{u^3}{3} + C_2$.

Substitute back $u = \cos x$: $-\frac{\cos^3 x}{3} + C_2$.

Combining the results:

$\int \sin^3 x \;dx = (-\cos x + C_1) - \left(-\frac{\cos^3 x}{3} + C_2\right)$

$-\cos x + \frac{\cos^3 x}{3} + (C_1 - C_2)$.

Let $C = C_1 - C_2$ be the constant of integration.

Therefore, $\int \sin^3 x \;dx = \mathbf{-\cos x + \frac{\cos^3 x}{3} + C}$.

To find the integral of $\sin^2 (2x + 5)$.

We use the trigonometric identity: $\mathbf{\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}}$.

Applying this identity with $\theta = 2x + 5$, we get:

$\sin^2 (2x + 5) = \frac{1 - \cos(2(2x + 5))}{2} = \frac{1 - \cos(4x + 10)}{2}$

Now, we integrate this expression:

$\int \sin^2 (2x + 5) \, dx = \int \frac{1 - \cos(4x + 10)}{2} \, dx$

$= \frac{1}{2} \int (1 - \cos(4x + 10)) \, dx$

$= \frac{1}{2} \left( \int 1 \, dx - \int \cos(4x + 10) \, dx \right)$

Integrating term by term:

$\int 1 \, dx = x$

For $\int \cos(4x + 10) \, dx$, let $u = 4x + 10$. Then $du = 4 \, dx$, so $dx = \frac{1}{4} du$.

$\int \cos(4x + 10) \, dx = \int \cos(u) \frac{1}{4} du = \frac{1}{4} \int \cos(u) \, du = \frac{1}{4} \sin(u) = \frac{1}{4} \sin(4x + 10)$

Substituting these back into the main integral, and adding the constant of integration $C$:

$\int \sin^2 (2x + 5) \, dx = \frac{1}{2} \left( x - \frac{1}{4} \sin(4x + 10) \right) + C$

Simplifying, the final integral is:

$\mathbf{\int \sin^2 (2x + 5) \, dx = \frac{x}{2} - \frac{1}{8} \sin(4x + 10) + C}$

To find the integral of $\sin(3x)\cos(4x)$.

We use the product-to-sum trigonometric identity: $\mathbf{2 \sin A \cos B = \sin(A+B) + \sin(A-B)}$.

Therefore, $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$.

Let $A = 3x$ and $B = 4x$. Applying the identity, we get:

$\sin(3x)\cos(4x) = \frac{1}{2} [\sin(3x+4x) + \sin(3x-4x)] $

$= \frac{1}{2} [\sin(7x) + \sin(-x)]$

Since $\sin(-x) = -\sin(x)$, the expression becomes:

$= \frac{1}{2} [\sin(7x) - \sin(x)]$

Now, we integrate the transformed expression:

$\int \sin(3x)\cos(4x) \, dx = \int \frac{1}{2} [\sin(7x) - \sin(x)] \, dx$

$= \frac{1}{2} \int (\sin(7x) - \sin(x)) \, dx$

$= \frac{1}{2} \left( \int \sin(7x) \, dx - \int \sin(x) \, dx \right)$

Integrating each term:

$\int \sin(7x) \, dx = -\frac{1}{7}\cos(7x)$

$\int \sin(x) \, dx = -\cos(x)$

Substituting these results back and adding the constant of integration $C$:

$\int \sin(3x)\cos(4x) \, dx = \frac{1}{2} \left( -\frac{1}{7}\cos(7x) - (-\cos(x)) \right) + C$

$= \frac{1}{2} \left( \cos(x) - \frac{1}{7}\cos(7x) \right) + C$

Simplifying, the final integral is:

$\mathbf{\int \sin(3x)\cos(4x) \, dx = \frac{1}{2}\cos(x) - \frac{1}{14}\cos(7x) + C}$

To find the integral of $\cos(2x)\cos(4x)\cos(6x)$.

We use the product-to-sum trigonometric identity: $\mathbf{2 \cos A \cos B = \cos(A+B) + \cos(A-B)}$.

We can rewrite the integrand as $\frac{1}{2} (2 \cos 4x \cos 2x) \cos 6x$.

Applying the identity to $2 \cos 4x \cos 2x$ (with $A=4x, B=2x$):

$2 \cos 4x \cos 2x = \cos(4x+2x) + \cos(4x-2x) = \cos(6x) + \cos(2x)$

So, the integrand becomes:

$\cos(2x)\cos(4x)\cos(6x) = \frac{1}{2} (\cos(6x) + \cos(2x)) \cos(6x)$

$= \frac{1}{2} (\cos^2(6x) + \cos(2x)\cos(6x))$

Now, we need to deal with $\cos^2(6x)$ and $\cos(2x)\cos(6x)$.

For $\cos^2(6x)$, we use the identity $\mathbf{\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}}$:

$\cos^2(6x) = \frac{1 + \cos(2 \times 6x)}{2} = \frac{1 + \cos(12x)}{2}$

For $\cos(2x)\cos(6x)$, we use the product-to-sum identity again (with $A=6x, B=2x$):

$\cos(2x)\cos(6x) = \frac{1}{2} (2 \cos 6x \cos 2x) = \frac{1}{2} (\cos(6x+2x) + \cos(6x-2x))$

$= \frac{1}{2} (\cos(8x) + \cos(4x))$

Substitute these back into the expression for the integrand:

$\cos(2x)\cos(4x)\cos(6x) = \frac{1}{2} \left( \frac{1 + \cos(12x)}{2} + \frac{1}{2} (\cos(8x) + \cos(4x)) \right)$

$= \frac{1}{2} \left( \frac{1}{2} + \frac{1}{2}\cos(12x) + \frac{1}{2}\cos(8x) + \frac{1}{2}\cos(4x) \right)$

$= \frac{1}{4} (1 + \cos(12x) + \cos(8x) + \cos(4x))$

Now, integrate the expression:

$\int \cos(2x)\cos(4x)\cos(6x) \, dx = \int \frac{1}{4} (1 + \cos(12x) + \cos(8x) + \cos(4x)) \, dx$

$= \frac{1}{4} \int (1 + \cos(12x) + \cos(8x) + \cos(4x)) \, dx$

$= \frac{1}{4} \left( \int 1 \, dx + \int \cos(12x) \, dx + \int \cos(8x) \, dx + \int \cos(4x) \, dx \right)$

Integrating each term:

$\int 1 \, dx = x$

$\int \cos(12x) \, dx = \frac{1}{12}\sin(12x)$

$\int \cos(8x) \, dx = \frac{1}{8}\sin(8x)$

$\int \cos(4x) \, dx = \frac{1}{4}\sin(4x)$

Substituting these results back and adding the constant of integration $C$:

$\int \cos(2x)\cos(4x)\cos(6x) \, dx = \frac{1}{4} \left( x + \frac{1}{12}\sin(12x) + \frac{1}{8}\sin(8x) + \frac{1}{4}\sin(4x) \right) + C$

Distributing the $\frac{1}{4}$, the final integral is:

$\mathbf{\int \cos(2x)\cos(4x)\cos(6x) \, dx = \frac{x}{4} + \frac{1}{48}\sin(12x) + \frac{1}{32}\sin(8x) + \frac{1}{16}\sin(4x) + C}$

To find the integral of $\sin^3 (2x + 1)$.

Let the integral be $I = \int \sin^3 (2x + 1) \, dx$.

We can rewrite $\sin^3 \theta$ as $\sin^2 \theta \cdot \sin \theta$.

So, $\sin^3 (2x + 1) = \sin^2 (2x + 1) \sin (2x + 1)$.

Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$, we get:

$\sin^3 (2x + 1) = (1 - \cos^2 (2x + 1)) \sin (2x + 1)$.

Now, let's substitute $u = 2x + 1$.

Then $du = 2 \, dx$, which means $dx = \frac{1}{2} \, du$.

The integral becomes:

$I = \int (1 - \cos^2 u) \sin u \left(\frac{1}{2} \, du\right)$

$I = \frac{1}{2} \int (1 - \cos^2 u) \sin u \, du$

Now, let's use another substitution. Let $v = \cos u$.

Then $dv = -\sin u \, du$, which means $\sin u \, du = -dv$.

The integral becomes:

$I = \frac{1}{2} \int (1 - v^2) (-dv)$

$I = -\frac{1}{2} \int (1 - v^2) \, dv$

Now, we can integrate with respect to $v$:

$I = -\frac{1}{2} \left( \int 1 \, dv - \int v^2 \, dv \right)$

$I = -\frac{1}{2} \left( v - \frac{v^3}{3} \right) + C_1$

Substitute back $v = \cos u$:

$I = -\frac{1}{2} \left( \cos u - \frac{\cos^3 u}{3} \right) + C_1$

$I = -\frac{1}{2}\cos u + \frac{1}{6}\cos^3 u + C_1$

Substitute back $u = 2x + 1$:

$I = -\frac{1}{2}\cos (2x + 1) + \frac{1}{6}\cos^3 (2x + 1) + C$

Where $C$ is the constant of integration.

Thus, the integral is:

$\mathbf{\int \sin^3 (2x + 1) \, dx = \frac{1}{6}\cos^3 (2x + 1) - \frac{1}{2}\cos (2x + 1) + C}$

To find the integral of $\sin^3 x \cos^3 x$.

We can rewrite the integrand by separating one term of $\cos x$ and using the identity $\mathbf{\cos^2 x = 1 - \sin^2 x}$.

$\sin^3 x \cos^3 x = \sin^3 x \cos^2 x \cos x$

$= \sin^3 x (1 - \sin^2 x) \cos x$

$= (\sin^3 x - \sin^5 x) \cos x$

Let the integral be $I = \int (\sin^3 x - \sin^5 x) \cos x \, dx$.

We use the method of substitution.

Let $\mathbf{u = \sin x}$.

Then, the differential $du$ is $\mathbf{du = \cos x \, dx}$.

Substituting $u$ and $du$ into the integral, we get:

$I = \int (u^3 - u^5) \, du$

Now, we integrate this polynomial with respect to $u$:

$I = \int u^3 \, du - \int u^5 \, du$

Performing the integration for each term:

$\int u^3 \, du = \frac{u^{3+1}}{3+1} = \frac{u^4}{4}$

$\int u^5 \, du = \frac{u^{5+1}}{5+1} = \frac{u^6}{6}$

So, the integral in terms of $u$ is:

$I = \frac{u^4}{4} - \frac{u^6}{6} + C$

where $C$ is the constant of integration.

Finally, substitute back $\mathbf{u = \sin x}$ to express the result in terms of $x$:

$I = \frac{(\sin x)^4}{4} - \frac{(\sin x)^6}{6} + C$

$I = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C$

Thus, the integral of $\sin^3 x \cos^3 x$ is:

$\mathbf{\int \sin^3 x \cos^3 x \, dx = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C}$

Alternate Solution:

We can also write the integrand using the identity $\mathbf{\sin(2x) = 2 \sin x \cos x}$.

$\sin^3 x \cos^3 x = (\sin x \cos x)^3 = \left(\frac{1}{2}\sin(2x)\right)^3 = \frac{1}{8}\sin^3(2x)$.

The integral becomes $I = \int \frac{1}{8}\sin^3(2x) \, dx = \frac{1}{8} \int \sin^3(2x) \, dx$.

Let $u = 2x$. Then $du = 2 \, dx$, so $dx = \frac{1}{2} \, du$.

$I = \frac{1}{8} \int \sin^3 u \left(\frac{1}{2} \, du\right) = \frac{1}{16} \int \sin^3 u \, du$.

To integrate $\sin^3 u$, we write it as $\sin u (1-\cos^2 u)$ and substitute $v = \cos u$, so $dv = -\sin u \, du$.

$\int \sin^3 u \, du = \int (1-\cos^2 u)\sin u \, du = \int (1-v^2)(-dv) = \int (v^2-1) \, dv$

$= \frac{v^3}{3} - v + C_1$.

Substituting back $v = \cos u$: $\frac{\cos^3 u}{3} - \cos u + C_1$.

Substituting back $u = 2x$:

$\int \sin^3(2x) \, dx = \frac{\cos^3(2x)}{3} - \cos(2x) + C_2$.

Therefore, the integral $I$ is:

$I = \frac{1}{16} \left(\frac{\cos^3(2x)}{3} - \cos(2x)\right) + C$

$\mathbf{I = \frac{\cos^3(2x)}{48} - \frac{\cos(2x)}{16} + C}$

To find the integral of $\sin x \sin 2x \sin 3x$.

We use the product-to-sum trigonometric identities. We can group the terms in different ways. Let's start by grouping $\sin x$ and $\sin 2x$.

Using the identity $\mathbf{2 \sin A \sin B = \cos(A-B) - \cos(A+B)}$, we have:

$2 \sin 2x \sin x = \cos(2x - x) - \cos(2x + x) = \cos x - \cos 3x$

So, $\sin x \sin 2x = \frac{1}{2}(\cos x - \cos 3x)$.

Now, multiply this result by $\sin 3x$:

$\sin x \sin 2x \sin 3x = \frac{1}{2}(\cos x - \cos 3x) \sin 3x$

$= \frac{1}{2} (\cos x \sin 3x - \cos 3x \sin 3x)$

We now need to transform the terms $\cos x \sin 3x$ and $\cos 3x \sin 3x$ using other identities.

For $\cos x \sin 3x$, use the identity $\mathbf{2 \cos A \sin B = \sin(A+B) - \sin(A-B)}$ with $A = x$ and $B = 3x$:

$2 \cos x \sin 3x = \sin(x + 3x) - \sin(x - 3x) = \sin(4x) - \sin(-2x)$

Since $\sin(-2x) = -\sin(2x)$, this becomes:

$= \sin(4x) + \sin(2x)$

So, $\cos x \sin 3x = \frac{1}{2}(\sin(4x) + \sin(2x))$.

For $\cos 3x \sin 3x$, use the identity $\mathbf{\sin 2\theta = 2 \sin \theta \cos \theta}$ with $\theta = 3x$:

$\cos 3x \sin 3x = \frac{1}{2} (2 \sin 3x \cos 3x) = \frac{1}{2}\sin(2 \times 3x) = \frac{1}{2}\sin(6x)$.

Substitute these back into the expression for the integrand:

$\sin x \sin 2x \sin 3x = \frac{1}{2} \left[ \frac{1}{2}(\sin(4x) + \sin(2x)) - \frac{1}{2}\sin(6x) \right]$

$= \frac{1}{4} [\sin(4x) + \sin(2x) - \sin(6x)]$

Rearranging the terms, we get:

$= \frac{1}{4} (\sin(2x) + \sin(4x) - \sin(6x))$

Now, we integrate this expression term by term:

$\int \sin x \sin 2x \sin 3x \, dx = \int \frac{1}{4} (\sin(2x) + \sin(4x) - \sin(6x)) \, dx$

$= \frac{1}{4} \int (\sin(2x) + \sin(4x) - \sin(6x)) \, dx$

$= \frac{1}{4} \left( \int \sin(2x) \, dx + \int \sin(4x) \, dx - \int \sin(6x) \, dx \right)$

Recall that $\int \sin(ax) \, dx = -\frac{1}{a}\cos(ax)$. Applying this to each term:

$\int \sin(2x) \, dx = -\frac{1}{2}\cos(2x)$

$\int \sin(4x) \, dx = -\frac{1}{4}\cos(4x)$

$\int \sin(6x) \, dx = -\frac{1}{6}\cos(6x)$

Substitute these results back into the integral expression and add the constant of integration $C$:

$\int \sin x \sin 2x \sin 3x \, dx = \frac{1}{4} \left( -\frac{1}{2}\cos(2x) - \frac{1}{4}\cos(4x) - (-\frac{1}{6}\cos(6x)) \right) + C$

$= \frac{1}{4} \left( -\frac{1}{2}\cos(2x) - \frac{1}{4}\cos(4x) + \frac{1}{6}\cos(6x) \right) + C$

Distributing the $\frac{1}{4}$, we get the final integral:

$\mathbf{\int \sin x \sin 2x \sin 3x \, dx = -\frac{1}{8}\cos(2x) - \frac{1}{16}\cos(4x) + \frac{1}{24}\cos(6x) + C}$

This can also be written as $\frac{1}{24}\cos(6x) - \frac{1}{8}\cos(2x) - \frac{1}{16}\cos(4x) + C$.

Alternate initial pairing:

We could have started by considering the product of $\sin x \sin 3x$. Using the identity $\mathbf{2 \sin A \sin B = \cos(A-B) - \cos(A+B)}$ with $A = 3x$ and $B = x$:

$2 \sin 3x \sin x = \cos(3x - x) - \cos(3x + x) = \cos(2x) - \cos(4x)$

So, $\sin x \sin 3x = \frac{1}{2}(\cos(2x) - \cos(4x))$.

Now, multiply this by $\sin 2x$:

$\sin x \sin 2x \sin 3x = \frac{1}{2}(\cos(2x) - \cos(4x)) \sin 2x$

$= \frac{1}{2} (\cos(2x)\sin 2x - \cos(4x)\sin 2x)$

Consider the terms $\cos(2x)\sin 2x$ and $\cos(4x)\sin 2x$ separately.

For $\cos(2x)\sin 2x$, use the identity $\mathbf{\sin 2\theta = 2 \sin \theta \cos \theta}$ with $\theta = 2x$:

$\cos(2x)\sin 2x = \frac{1}{2} (2 \sin 2x \cos 2x) = \frac{1}{2}\sin(2 \times 2x) = \frac{1}{2}\sin(4x)$.

For $\cos(4x)\sin 2x$, use the identity $\mathbf{2 \cos A \sin B = \sin(A+B) - \sin(A-B)}$ with $A = 4x$ and $B = 2x$:

$2 \cos 4x \sin 2x = \sin(4x + 2x) - \sin(4x - 2x) = \sin(6x) - \sin(2x)$

So, $\cos(4x)\sin 2x = \frac{1}{2}(\sin(6x) - \sin(2x))$.

Substitute these back into the expression for the integrand:

$\sin x \sin 2x \sin 3x = \frac{1}{2} \left[ \frac{1}{2}\sin(4x) - \frac{1}{2}(\sin(6x) - \sin(2x)) \right]$

$= \frac{1}{4} (\sin(4x) - \sin(6x) + \sin(2x))$

$= \frac{1}{4} (\sin(2x) + \sin(4x) - \sin(6x))$

This confirms that starting with a different pairing leads to the same expression before integration.

The integration proceeds as shown previously.

The final integral is $\mathbf{-\frac{1}{8}\cos(2x) - \frac{1}{16}\cos(4x) + \frac{1}{24}\cos(6x) + C}$.

To find the integral of $\sin(4x) \sin(8x)$.

We use the product-to-sum trigonometric identity: $\mathbf{2 \sin A \sin B = \cos(A-B) - \cos(A+B)}$.

Let $A = 4x$ and $B = 8x$. Then the given function can be written as:

$\sin(4x) \sin(8x) = \frac{1}{2} [2 \sin(4x) \sin(8x)]$

Applying the identity:

$= \frac{1}{2} [\cos(4x - 8x) - \cos(4x + 8x)]$

$= \frac{1}{2} [\cos(-4x) - \cos(12x)]$

Since $\cos(-\theta) = \cos(\theta)$, we have:

$= \frac{1}{2} [\cos(4x) - \cos(12x)]$

Now, we integrate this transformed expression:

$\int \sin(4x) \sin(8x) \, dx = \int \frac{1}{2} [\cos(4x) - \cos(12x)] \, dx$

$= \frac{1}{2} \int (\cos(4x) - \cos(12x)) \, dx$

$= \frac{1}{2} \left( \int \cos(4x) \, dx - \int \cos(12x) \, dx \right)$

Integrating each term using the formula $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax)$:

$\int \cos(4x) \, dx = \frac{1}{4}\sin(4x)$

$\int \cos(12x) \, dx = \frac{1}{12}\sin(12x)$

Substitute these results back into the integral expression and add the constant of integration $C$:

$\int \sin(4x) \sin(8x) \, dx = \frac{1}{2} \left( \frac{1}{4}\sin(4x) - \frac{1}{12}\sin(12x) \right) + C$

Distributing the $\frac{1}{2}$, we get the final integral:

$\mathbf{\int \sin(4x) \sin(8x) \, dx = \frac{1}{8}\sin(4x) - \frac{1}{24}\sin(12x) + C}$

To find the integral of $\frac{1 - \cos x}{1 + \cos x}$.

We use the half-angle trigonometric identities:

$\mathbf{1 - \cos x = 2 \sin^2 \frac{x}{2}}$

$\mathbf{1 + \cos x = 2 \cos^2 \frac{x}{2}}$

Substitute these identities into the integrand:

$\frac{1 - \cos x}{1 + \cos x} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}}$

$= \frac{\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}}$

Recall that $\frac{\sin \theta}{\cos \theta} = \tan \theta$. So, $\frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$.

Thus, the integrand simplifies to:

$= \tan^2 \frac{x}{2}$

Now, we need to integrate $\tan^2 \frac{x}{2}$. We use the identity $\mathbf{\tan^2 \theta = \sec^2 \theta - 1}$.

So, $\tan^2 \frac{x}{2} = \sec^2 \frac{x}{2} - 1$.

The integral becomes:

$\int \frac{1 - \cos x}{1 + \cos x} \, dx = \int \left(\sec^2 \frac{x}{2} - 1\right) \, dx$

$= \int \sec^2 \frac{x}{2} \, dx - \int 1 \, dx$

Integrate the first term $\int \sec^2 \frac{x}{2} \, dx$. Let $u = \frac{x}{2}$. Then $du = \frac{1}{2} \, dx$, which means $dx = 2 \, du$.

$\int \sec^2 \frac{x}{2} \, dx = \int \sec^2 u \, (2 \, du) = 2 \int \sec^2 u \, du$

We know that $\int \sec^2 u \, du = \tan u$.

So, $2 \int \sec^2 u \, du = 2 \tan u$.

Substitute back $u = \frac{x}{2}$:

$= 2 \tan \frac{x}{2}$.

Integrate the second term $\int 1 \, dx = x$.

Combining the results and adding the constant of integration $C$:

$\int \frac{1 - \cos x}{1 + \cos x} \, dx = 2 \tan \frac{x}{2} - x + C$

Thus, the integral is:

$\mathbf{\int \frac{1 - \cos x}{1 + \cos x} \, dx = 2 \tan \frac{x}{2} - x + C}$

To find the integral of $\frac{\cos x}{1 + \cos x}$.

We can rewrite the integrand by adding and subtracting 1 in the numerator:

$\frac{\cos x}{1 + \cos x} = \frac{\cos x + 1 - 1}{1 + \cos x}$

$= \frac{1 + \cos x}{1 + \cos x} - \frac{1}{1 + \cos x}$

$= 1 - \frac{1}{1 + \cos x}$

Now, we integrate this expression:

$\int \frac{\cos x}{1 + \cos x} \, dx = \int \left(1 - \frac{1}{1 + \cos x}\right) \, dx$

$= \int 1 \, dx - \int \frac{1}{1 + \cos x} \, dx$

We know that $\int 1 \, dx = x$.

For the second integral, we use the half-angle identity: $\mathbf{1 + \cos x = 2 \cos^2 \frac{x}{2}}$.

So, $\int \frac{1}{1 + \cos x} \, dx = \int \frac{1}{2 \cos^2 \frac{x}{2}} \, dx$

$= \frac{1}{2} \int \frac{1}{\cos^2 \frac{x}{2}} \, dx$

Recall that $\frac{1}{\cos \theta} = \sec \theta$, so $\frac{1}{\cos^2 \theta} = \sec^2 \theta$.

$= \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$

To evaluate $\int \sec^2 \frac{x}{2} \, dx$, let $u = \frac{x}{2}$.

Then, $du = \frac{1}{2} \, dx$, which implies $dx = 2 \, du$.

The integral becomes:

$\frac{1}{2} \int \sec^2 u \, (2 \, du) = \int \sec^2 u \, du$

The integral of $\sec^2 u$ is $\tan u$.

$= \tan u$

Substitute back $u = \frac{x}{2}$:

$= \tan \frac{x}{2}$

Combining the results of both parts of the integral and adding the constant of integration $C$:

$\int \frac{\cos x}{1 + \cos x} \, dx = x - \tan \frac{x}{2} + C$

Thus, the integral is:

$\mathbf{\int \frac{\cos x}{1 + \cos x} \, dx = x - \tan \frac{x}{2} + C}$

To find the integral of $\sin^4 x$.

We can rewrite $\sin^4 x$ as $(\sin^2 x)^2$.

Using the identity $\mathbf{\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}}$, with $\theta = x$, we have:

$\sin^2 x = \frac{1 - \cos(2x)}{2}$

So, $\sin^4 x = \left(\frac{1 - \cos(2x)}{2}\right)^2$

$= \frac{(1 - \cos(2x))^2}{4}$

$= \frac{1 - 2\cos(2x) + \cos^2(2x)}{4}$

Now, we need to deal with the $\cos^2(2x)$ term. Using the identity $\mathbf{\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}}$, with $\theta = 2x$, we have:

$\cos^2(2x) = \frac{1 + \cos(2 \times 2x)}{2} = \frac{1 + \cos(4x)}{2}$

Substitute this back into the expression for $\sin^4 x$:

$\sin^4 x = \frac{1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}}{4}$

$= \frac{1}{4} \left( 1 - 2\cos(2x) + \frac{1}{2} + \frac{1}{2}\cos(4x) \right)$

Combine the constant terms:

$= \frac{1}{4} \left( \left(1 + \frac{1}{2}\right) - 2\cos(2x) + \frac{1}{2}\cos(4x) \right)$

$= \frac{1}{4} \left( \frac{3}{2} - 2\cos(2x) + \frac{1}{2}\cos(4x) \right)$

Distribute the $\frac{1}{4}$:

$= \frac{3}{8} - \frac{2}{4}\cos(2x) + \frac{1}{8}\cos(4x)$

$= \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)$

Now, we integrate this simplified expression term by term:

$\int \sin^4 x \, dx = \int \left(\frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\right) \, dx$

$= \int \frac{3}{8} \, dx - \int \frac{1}{2}\cos(2x) \, dx + \int \frac{1}{8}\cos(4x) \, dx$

$= \frac{3}{8} \int 1 \, dx - \frac{1}{2} \int \cos(2x) \, dx + \frac{1}{8} \int \cos(4x) \, dx$

Recall that $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax)$. Integrating each term:

$\int 1 \, dx = x$

$\int \cos(2x) \, dx = \frac{1}{2}\sin(2x)$

$\int \cos(4x) \, dx = \frac{1}{4}\sin(4x)$

Substitute these results back into the integral expression and add the constant of integration $C$:

$\int \sin^4 x \, dx = \frac{3}{8}x - \frac{1}{2}\left(\frac{1}{2}\sin(2x)\right) + \frac{1}{8}\left(\frac{1}{4}\sin(4x)\right) + C$

$= \frac{3}{8}x - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C$

Thus, the integral of $\sin^4 x$ is:

$\mathbf{\int \sin^4 x \, dx = \frac{3}{8}x - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C}$

To find the integral of $\cos^4 (2x)$.

We can rewrite $\cos^4 (2x)$ as $(\cos^2 (2x))^2$.

Using the trigonometric identity: $\mathbf{\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}}$.

Let $\theta = 2x$. Then $\cos^2 (2x) = \frac{1 + \cos(2 \times 2x)}{2} = \frac{1 + \cos(4x)}{2}$.

Substitute this back into the expression for $\cos^4 (2x)$:

$\cos^4 (2x) = \left(\frac{1 + \cos(4x)}{2}\right)^2$

$= \frac{(1 + \cos(4x))^2}{4}$

Expand the numerator:

$= \frac{1 + 2\cos(4x) + \cos^2(4x)}{4}$

Now we need to simplify the $\cos^2(4x)$ term using the same identity $\mathbf{\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}}$.

Let $\theta = 4x$. Then $\cos^2 (4x) = \frac{1 + \cos(2 \times 4x)}{2} = \frac{1 + \cos(8x)}{2}$.

Substitute this back into the expression for $\cos^4 (2x)$:

$\cos^4 (2x) = \frac{1 + 2\cos(4x) + \frac{1 + \cos(8x)}{2}}{4}$

$= \frac{1}{4} \left( 1 + 2\cos(4x) + \frac{1}{2} + \frac{1}{2}\cos(8x) \right)$

Combine the constant terms:

$= \frac{1}{4} \left( \frac{3}{2} + 2\cos(4x) + \frac{1}{2}\cos(8x) \right)$

Distribute the $\frac{1}{4}$:

$= \frac{3}{8} + \frac{2}{4}\cos(4x) + \frac{1}{8}\cos(8x)$

$= \frac{3}{8} + \frac{1}{2}\cos(4x) + \frac{1}{8}\cos(8x)$

Now, we integrate this simplified expression term by term:

$\int \cos^4 (2x) \, dx = \int \left(\frac{3}{8} + \frac{1}{2}\cos(4x) + \frac{1}{8}\cos(8x)\right) \, dx$

$= \int \frac{3}{8} \, dx + \int \frac{1}{2}\cos(4x) \, dx + \int \frac{1}{8}\cos(8x) \, dx$

$= \frac{3}{8} \int 1 \, dx + \frac{1}{2} \int \cos(4x) \, dx + \frac{1}{8} \int \cos(8x) \, dx$

We use the integral formula $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax)$.

$\int 1 \, dx = x$

$\int \cos(4x) \, dx = \frac{1}{4}\sin(4x)$

$\int \cos(8x) \, dx = \frac{1}{8}\sin(8x)$

Substitute these results back into the integral expression and add the constant of integration $C$:

$\int \cos^4 (2x) \, dx = \frac{3}{8}x + \frac{1}{2}\left(\frac{1}{4}\sin(4x)\right) + \frac{1}{8}\left(\frac{1}{8}\sin(8x)\right) + C$

$= \frac{3}{8}x + \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + C$

Thus, the integral of $\cos^4 (2x)$ is:

$\mathbf{\int \cos^4 (2x) \, dx = \frac{3}{8}x + \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + C}$

To find the integral of $\frac{\sin^2 x}{1 + \cos x}$.

We can simplify the integrand using trigonometric identities.

Recall the identity $\mathbf{\sin^2 x + \cos^2 x = 1}$, which implies $\mathbf{\sin^2 x = 1 - \cos^2 x}$.

Substitute this into the numerator:

$\frac{\sin^2 x}{1 + \cos x} = \frac{1 - \cos^2 x}{1 + \cos x}$

The numerator is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a=1$ and $b=\cos x$.

$1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$

So, the integrand becomes:

$\frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x}$

Assuming $1 + \cos x \neq 0$, we can cancel the $(1 + \cos x)$ term from the numerator and denominator:

$\frac{(1 - \cos x)\cancel{(1 + \cos x)}}{\cancel{1 + \cos x}} = 1 - \cos x$

The original expression simplifies to $1 - \cos x$.

Now, we integrate the simplified expression:

$\int \frac{\sin^2 x}{1 + \cos x} \, dx = \int (1 - \cos x) \, dx$

$= \int 1 \, dx - \int \cos x \, dx$

Integrating each term:

$\int 1 \, dx = x$

$\int \cos x \, dx = \sin x$

Combining the results and adding the constant of integration $C$:

$\int (1 - \cos x) \, dx = x - \sin x + C$

Thus, the integral is:

$\mathbf{\int \frac{\sin^2 x}{1 + \cos x} \, dx = x - \sin x + C}$

To find the integral of $\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}$.

We use the double-angle identity for cosine: $\mathbf{\cos 2\theta = 2\cos^2 \theta - 1}$.

Substitute this into the numerator:

$\cos 2x = 2\cos^2 x - 1$

$\cos 2\alpha = 2\cos^2 \alpha - 1$

So, the numerator $\cos 2x - \cos 2\alpha$ becomes:

$(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)$

$= 2\cos^2 x - 1 - 2\cos^2 \alpha + 1$

$= 2\cos^2 x - 2\cos^2 \alpha$

$= 2(\cos^2 x - \cos^2 \alpha)$

Now, substitute this back into the integrand:

$\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} = \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha}$

The term in the parenthesis in the numerator is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a=\cos x$ and $b=\cos \alpha$.

$\cos^2 x - \cos^2 \alpha = (\cos x - \cos \alpha)(\cos x + \cos \alpha)$

So, the integrand becomes:

$\frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha}$

Assuming $\cos x - \cos \alpha \neq 0$, we can cancel the $(\cos x - \cos \alpha)$ term from the numerator and denominator:

$\frac{2\cancel{(\cos x - \cos \alpha)}(\cos x + \cos \alpha)}{\cancel{\cos x - \cos \alpha}} = 2(\cos x + \cos \alpha)$

The integrand simplifies to $2(\cos x + \cos \alpha)$. Note that $\alpha$ is a constant, so $\cos \alpha$ is also a constant.

Now, we integrate this simplified expression:

$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx = \int 2(\cos x + \cos \alpha) \, dx$

$= 2 \int (\cos x + \cos \alpha) \, dx$

$= 2 \left( \int \cos x \, dx + \int \cos \alpha \, dx \right)$

Integrating each term:

$\int \cos x \, dx = \sin x$

Since $\cos \alpha$ is a constant with respect to $x$, $\int \cos \alpha \, dx = (\cos \alpha) \int 1 \, dx = (\cos \alpha) x$.

Combining the results and adding the constant of integration $C$:

$\int 2(\cos x + \cos \alpha) \, dx = 2 (\sin x + x \cos \alpha) + C$

$= 2\sin x + 2x\cos \alpha + C$

Thus, the integral is:

$\mathbf{\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx = 2\sin x + 2x\cos \alpha + C}$

To find the integral of $\frac{\cos x - \sin x}{1 + \sin 2x}$.

We can simplify the denominator using trigonometric identities.

Recall that $\mathbf{1 = \sin^2 x + \cos^2 x}$ and $\mathbf{\sin 2x = 2 \sin x \cos x}$.

So, the denominator $1 + \sin 2x$ can be written as:

$1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$

This is in the form of a perfect square: $(a+b)^2 = a^2 + b^2 + 2ab$.

$= (\sin x + \cos x)^2$

Substitute this back into the integrand:

$\frac{\cos x - \sin x}{1 + \sin 2x} = \frac{\cos x - \sin x}{(\sin x + \cos x)^2}$

Let the integral be $I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx$.

We use the method of substitution.

Let $\mathbf{u = \sin x + \cos x}$.

Then, the differential $du$ is the derivative of $u$ with respect to $x$ multiplied by $dx$.

$\frac{du}{dx} = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$

So, the differential $\mathbf{du = (\cos x - \sin x) \, dx}$.

Notice that the numerator of the integrand is exactly $du$. The denominator is $u^2$.

Substituting $u$ and $du$ into the integral, we get:

$I = \int \frac{du}{u^2}$

$I = \int u^{-2} \, du$

Now, we integrate $u^{-2}$ using the power rule for integration $\int u^n \, du = \frac{u^{n+1}}{n+1}$ (for $n \neq -1$).

Here, $n = -2$.

$I = \frac{u^{-2+1}}{-2+1} + C$

$I = \frac{u^{-1}}{-1} + C$

$I = -\frac{1}{u} + C$

where $C$ is the constant of integration.

Finally, substitute back $\mathbf{u = \sin x + \cos x}$ to express the result in terms of $x$:

$I = -\frac{1}{\sin x + \cos x} + C$

Thus, the integral is:

$\mathbf{\int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx = -\frac{1}{\sin x + \cos x} + C}$

To find the integral of $\tan^3(2x) \sec(2x)$.

Let the integral be $I = \int \tan^3(2x) \sec(2x) \, dx$.

We can rewrite the integrand to facilitate substitution.

$\tan^3(2x) \sec(2x) = \tan^2(2x) \cdot \tan(2x) \sec(2x)$

Using the identity $\mathbf{\tan^2 \theta = \sec^2 \theta - 1}$, with $\theta = 2x$, we have:

$\tan^2(2x) = \sec^2(2x) - 1$

Substitute this back into the integrand:

$I = \int (\sec^2(2x) - 1) \tan(2x) \sec(2x) \, dx$

We use the method of substitution.

Let $\mathbf{u = \sec(2x)}$.

Then, the differential $du$ is the derivative of $u$ with respect to $x$ multiplied by $dx$.

$\frac{du}{dx} = \frac{d}{dx}(\sec(2x))$

Using the chain rule, $\frac{d}{dx}(\sec f(x)) = \sec f(x) \tan f(x) f'(x)$. Here $f(x) = 2x$, so $f'(x) = 2$.

$\frac{du}{dx} = \sec(2x) \tan(2x) \cdot 2 = 2 \sec(2x) \tan(2x)$

So, the differential $\mathbf{du = 2 \sec(2x) \tan(2x) \, dx}$, which means $\mathbf{\sec(2x) \tan(2x) \, dx = \frac{1}{2} \, du}$.

Substitute $u$ and $du$ into the integral. The term $\sec^2(2x)$ becomes $u^2$, and the term $\tan(2x)\sec(2x)\,dx$ becomes $\frac{1}{2}\,du$.

$I = \int (u^2 - 1) \left(\frac{1}{2} \, du\right)$

$I = \frac{1}{2} \int (u^2 - 1) \, du$

Now, we integrate this polynomial with respect to $u$:

$I = \frac{1}{2} \left( \int u^2 \, du - \int 1 \, du \right)$

Performing the integration for each term:

$\int u^2 \, du = \frac{u^{2+1}}{2+1} = \frac{u^3}{3}$

$\int 1 \, du = u$

So, the integral in terms of $u$ is:

$I = \frac{1}{2} \left( \frac{u^3}{3} - u \right) + C$

$I = \frac{u^3}{6} - \frac{u}{2} + C$

where $C$ is the constant of integration.

Finally, substitute back $\mathbf{u = \sec(2x)}$ to express the result in terms of $x$:

$I = \frac{(\sec(2x))^3}{6} - \frac{\sec(2x)}{2} + C$

$I = \frac{\sec^3(2x)}{6} - \frac{\sec(2x)}{2} + C$

Thus, the integral is:

$\mathbf{\int \tan^3(2x) \sec(2x) \, dx = \frac{\sec^3(2x)}{6} - \frac{\sec(2x)}{2} + C}$

To find the integral of $\tan^4 x$.

We can rewrite $\tan^4 x$ as $\tan^2 x \cdot \tan^2 x$.

Using the identity $\mathbf{\tan^2 x = \sec^2 x - 1}$, we substitute one $\tan^2 x$ term:

$\tan^4 x = \tan^2 x (\sec^2 x - 1)$

$= \tan^2 x \sec^2 x - \tan^2 x$

Now, we integrate the expression term by term:

$\int \tan^4 x \, dx = \int (\tan^2 x \sec^2 x - \tan^2 x) \, dx$

$= \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx$

Consider the first integral, $\int \tan^2 x \sec^2 x \, dx$.

Let $\mathbf{u = \tan x}$.

Then, the differential $du$ is $\mathbf{du = \sec^2 x \, dx}$.

Substituting $u$ and $du$, the integral becomes:

$\int u^2 \, du = \frac{u^{2+1}}{2+1} + C_1 = \frac{u^3}{3} + C_1$

Substitute back $u = \tan x$:

$= \frac{\tan^3 x}{3} + C_1$

Consider the second integral, $\int \tan^2 x \, dx$.

Using the identity $\mathbf{\tan^2 x = \sec^2 x - 1}$ again:

$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx$

$= \int \sec^2 x \, dx - \int 1 \, dx$

We know that $\int \sec^2 x \, dx = \tan x$ and $\int 1 \, dx = x$.

So, $\int \tan^2 x \, dx = \tan x - x + C_2$

Combine the results of the two integrals and add the constant of integration $C$ ($= C_1 - C_2$):

$\int \tan^4 x \, dx = \left(\frac{\tan^3 x}{3}\right) - (\tan x - x) + C$

$= \frac{\tan^3 x}{3} - \tan x + x + C$

Thus, the integral of $\tan^4 x$ is:

$\mathbf{\int \tan^4 x \, dx = \frac{\tan^3 x}{3} - \tan x + x + C}$

To find the integral of $\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x}$.

We can simplify the integrand by splitting the fraction:

$\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} = \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x}$

Simplify the first term:

$\frac{\sin^3 x}{\sin^2 x \cos^2 x} = \frac{\sin x}{\cos^2 x}$

Using the identities $\frac{1}{\cos x} = \sec x$ and $\frac{\sin x}{\cos x} = \tan x$, we can write this as:

$= \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x$

Simplify the second term:

$\frac{\cos^3 x}{\sin^2 x \cos^2 x} = \frac{\cos x}{\sin^2 x}$

Using the identities $\frac{1}{\sin x} = \text{cosec } x$ and $\frac{\cos x}{\sin x} = \cot x$, we can write this as:

$= \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \cot x \text{ cosec } x$

So the original integrand simplifies to:

$\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} = \tan x \sec x + \cot x \text{ cosec } x$

Now, we integrate the simplified expression term by term:

$\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx = \int (\tan x \sec x + \cot x \text{ cosec } x) \, dx$

$= \int \tan x \sec x \, dx + \int \cot x \text{ cosec } x \, dx$

We use the standard integral formulas:

$\mathbf{\int \sec x \tan x \, dx = \sec x + C_1}$

$\mathbf{\int \text{cosec } x \cot x \, dx = -\text{cosec } x + C_2}$

Combining the results and adding the constant of integration $C = C_1 + C_2$:

$\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx = \sec x - \text{cosec } x + C$

Thus, the integral is:

$\mathbf{\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx = \sec x - \text{cosec } x + C}$

To find the integral of $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$.

We can simplify the numerator using trigonometric identities.

Recall the double-angle identity for cosine in terms of sine: $\mathbf{\cos 2x = 1 - 2\sin^2 x}$.

Substitute this into the numerator:

$\cos 2x + 2\sin^2 x = (1 - 2\sin^2 x) + 2\sin^2 x$

$= 1 - 2\sin^2 x + 2\sin^2 x$

$= 1$

Substitute the simplified numerator back into the integrand:

$\frac{\cos 2x + 2\sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}$

Using the identity $\mathbf{\frac{1}{\cos x} = \sec x}$, we have $\frac{1}{\cos^2 x} = \sec^2 x$.

The integrand simplifies to $\sec^2 x$.

Now, we integrate the simplified expression:

$\int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx = \int \sec^2 x \, dx$

We know that the integral of $\sec^2 x$ is $\tan x$.

$\int \sec^2 x \, dx = \tan x + C$

where $C$ is the constant of integration.

Thus, the integral is:

$\mathbf{\int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx = \tan x + C}$

To find the integral of $\frac{1}{\sin x \cos^3 x}$.

We can rewrite the integrand by multiplying the numerator and denominator by $\sec^4 x$.

$\frac{1}{\sin x \cos^3 x} = \frac{1}{\sin x \cos^3 x} \times \frac{\sec^4 x}{\sec^4 x}$

$= \frac{\sec^4 x}{\sin x \cos^3 x \cdot \frac{1}{\cos^4 x}}$

$= \frac{\sec^4 x}{\frac{\sin x}{\cos x}}$

Using the identity $\frac{\sin x}{\cos x} = \tan x$, we get:

$= \frac{\sec^4 x}{\tan x}$

Now, we can rewrite $\sec^4 x$ as $\sec^2 x \cdot \sec^2 x$.

Using the identity $\mathbf{\sec^2 x = 1 + \tan^2 x}$, we have $\sec^4 x = (1 + \tan^2 x)\sec^2 x$.

So the integrand becomes:

$\frac{(1 + \tan^2 x)\sec^2 x}{\tan x}$

Let the integral be $I = \int \frac{(1 + \tan^2 x)\sec^2 x}{\tan x} \, dx$.

We use the method of substitution.

Let $\mathbf{u = \tan x}$.

Then, the differential $du$ is $\mathbf{du = \sec^2 x \, dx}$.

Substituting $u$ and $du$ into the integral, we get:

$I = \int \frac{1 + u^2}{u} \, du$

We can split the fraction:

$I = \int \left(\frac{1}{u} + \frac{u^2}{u}\right) \, du$

$I = \int \left(\frac{1}{u} + u\right) \, du$

Now, we integrate term by term with respect to $u$:

$I = \int \frac{1}{u} \, du + \int u \, du$

The integrals are:

$\int \frac{1}{u} \, du = \ln|u|$

$\int u \, du = \frac{u^{1+1}}{1+1} = \frac{u^2}{2}$

So, the integral in terms of $u$ is:

$I = \ln|u| + \frac{u^2}{2} + C$

where $C$ is the constant of integration.

Finally, substitute back $\mathbf{u = \tan x}$ to express the result in terms of $x$:

$I = \ln|\tan x| + \frac{(\tan x)^2}{2} + C$

$I = \ln|\tan x| + \frac{\tan^2 x}{2} + C$

Thus, the integral is:

$\mathbf{\int \frac{1}{\sin x \cos^3 x} \, dx = \ln|\tan x| + \frac{\tan^2 x}{2} + C}$

To find the integral of $\frac{\cos 2x}{(\cos x + \sin x)^2}$.

We can simplify the integrand using trigonometric identities.

Recall the double-angle identity for cosine: $\mathbf{\cos 2x = \cos^2 x - \sin^2 x}$.

Substitute this into the numerator:

$\frac{\cos 2x}{(\cos x + \sin x)^2} = \frac{\cos^2 x - \sin^2 x}{(\cos x + \sin x)^2}$

The numerator is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a=\cos x$ and $b=\sin x$.

$\cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$

So, the integrand becomes:

$\frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2}$

Assuming $\cos x + \sin x \neq 0$, we can cancel one factor of $(\cos x + \sin x)$ from the numerator and denominator:

$\frac{(\cos x - \sin x)\cancel{(\cos x + \sin x)}}{(\cos x + \sin x)\cancel{^2}} = \frac{\cos x - \sin x}{\cos x + \sin x}$

The integrand simplifies to $\frac{\cos x - \sin x}{\cos x + \sin x}$.

Let the integral be $I = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$.

We use the method of substitution.

Let $\mathbf{u = \cos x + \sin x}$.

Then, the differential $du$ is the derivative of $u$ with respect to $x$ multiplied by $dx$.

$\frac{du}{dx} = \frac{d}{dx}(\cos x + \sin x) = -\sin x + \cos x = \cos x - \sin x$

So, the differential $\mathbf{du = (\cos x - \sin x) \, dx}$.

Notice that the numerator of the integrand is exactly $du$. The denominator is $u$.

Substituting $u$ and $du$ into the integral, we get:

$I = \int \frac{du}{u}$

We know that $\int \frac{1}{u} \, du = \ln|u|$.

$I = \ln|u| + C$

where $C$ is the constant of integration.

Finally, substitute back $\mathbf{u = \cos x + \sin x}$ to express the result in terms of $x$:

$I = \ln|\cos x + \sin x| + C$

Thus, the integral is:

$\mathbf{\int \frac{\cos 2x}{(\cos x + \sin x)^2} \, dx = \ln|\cos x + \sin x| + C}$

To find the integral of $\sin^{-1}(\cos x)$.

We can simplify the integrand using the identity relating sine and cosine functions.

Recall that $\mathbf{\cos x = \sin\left(\frac{\pi}{2} - x\right)}$.

Substitute this into the argument of the inverse sine function:

$\sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right)$

For the identity $\sin^{-1}(\sin \theta) = \theta$ to hold, we typically need $\theta$ to be in the principal range of $\sin^{-1} u$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Assuming that $x$ is such that $\frac{\pi}{2} - x$ is within this range (e.g., $0 \leq x \leq \pi$), then $\sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x$.

So, the integrand simplifies to $\frac{\pi}{2} - x$, under appropriate conditions on $x$.

Now, we integrate the simplified expression:

$\int \sin^{-1}(\cos x) \, dx = \int \left(\frac{\pi}{2} - x\right) \, dx$

$= \int \frac{\pi}{2} \, dx - \int x \, dx$

Integrating each term:

$\int \frac{\pi}{2} \, dx = \frac{\pi}{2} \int 1 \, dx = \frac{\pi}{2}x$

$\int x \, dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}$

Combining the results and adding the constant of integration $C$:

$\int \left(\frac{\pi}{2} - x\right) \, dx = \frac{\pi}{2}x - \frac{x^2}{2} + C$

Thus, the integral is:

$\mathbf{\int \sin^{-1}(\cos x) \, dx = \frac{\pi}{2}x - \frac{x^2}{2} + C}$

Note: This solution assumes that the domain of $x$ is restricted such that $\frac{\pi}{2} - x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. This corresponds to $0 \leq x \leq \pi$. If $x$ is outside this range, $\sin^{-1}(\cos x)$ would be equal to something other than $\frac{\pi}{2} - x$, and the integral would need to be evaluated piecewise.

To find the integral of $\frac{1}{\cos (x - a) \cos (x - b)}$.

Let the integral be $I = \int \frac{1}{\cos (x - a) \cos (x - b)} \, dx$.

We multiply the numerator and denominator by a constant trigonometric term involving $(a-b)$. Consider $\sin(a-b)$.

We can write $\sin(a-b) = \sin((x-b) - (x-a))$.

Multiply the integrand by $\frac{\sin(a-b)}{\sin(a-b)}$ (assuming $\sin(a-b) \neq 0$, i.e., $a-b \neq n\pi$).

$I = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos (x - a) \cos (x - b)} \, dx$

$I = \frac{1}{\sin(a-b)} \int \frac{\sin((x-b) - (x-a))}{\cos (x - a) \cos (x - b)} \, dx$

Using the identity $\mathbf{\sin(A - B) = \sin A \cos B - \cos A \sin B}$ with $A = x-b$ and $B = x-a$ in the numerator:

$\sin((x-b) - (x-a)) = \sin(x-b)\cos(x-a) - \cos(x-b)\sin(x-a)$

Substitute this into the integral:

$I = \frac{1}{\sin(a-b)} \int \frac{\sin(x-b)\cos(x-a) - \cos(x-b)\sin(x-a)}{\cos (x - a) \cos (x - b)} \, dx$

Now, split the fraction into two terms:

$I = \frac{1}{\sin(a-b)} \int \left( \frac{\sin(x-b)\cos(x-a)}{\cos (x - a) \cos (x - b)} - \frac{\cos(x-b)\sin(x-a)}{\cos (x - a) \cos (x - b)} \right) \, dx$

Cancel the common terms in each fraction:

$I = \frac{1}{\sin(a-b)} \int \left( \frac{\sin(x-b)}{\cos (x - b)} - \frac{\sin(x-a)}{\cos (x - a)} \right) \, dx$

Using the identity $\mathbf{\frac{\sin \theta}{\cos \theta} = \tan \theta}$:

$I = \frac{1}{\sin(a-b)} \int (\tan(x-b) - \tan(x-a)) \, dx$

Integrate each term separately:

$I = \frac{1}{\sin(a-b)} \left( \int \tan(x-b) \, dx - \int \tan(x-a) \, dx \right)$

Using the standard integral $\mathbf{\int \tan \theta \, d\theta = \ln|\sec \theta| + C}$ or $\mathbf{-\ln|\cos \theta| + C}$:

$\int \tan(x-b) \, dx = -\ln|\cos(x-b)| + C_1$

$\int \tan(x-a) \, dx = -\ln|\cos(x-a)| + C_2$

Substitute these results back into the expression for $I$ and combine the constants of integration:

$I = \frac{1}{\sin(a-b)} \left( -\ln|\cos(x-b)| - (-\ln|\cos(x-a)|) \right) + C$

$I = \frac{1}{\sin(a-b)} \left( \ln|\cos(x-a)| - \ln|\cos(x-b)| \right) + C$

Using the logarithm property $\mathbf{\ln P - \ln Q = \ln \frac{P}{Q}}$:

$I = \frac{1}{\sin(a-b)} \ln \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C$

where $C$ is the constant of integration.

Thus, the integral is:

$\mathbf{\int \frac{1}{\cos (x − a) \cos (x − b)} \, dx = \frac{1}{\sin(a-b)} \ln \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C}$

Note: This solution is valid when $\sin(a-b) \neq 0$, i.e., $a-b \neq n\pi$ for any integer $n$. Also, the integral is defined where $\cos(x-a) \neq 0$ and $\cos(x-b) \neq 0$.

To find the integral of $\frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}$.

We can simplify the integrand by splitting the fraction into two parts:

$\frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} = \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x}$

Simplify the first term:

$\frac{\sin^2 x}{\sin^2 x \cos^2 x} = \frac{\cancel{\sin^2 x}}{\cancel{\sin^2 x} \cos^2 x} = \frac{1}{\cos^2 x}$

Using the identity $\mathbf{\frac{1}{\cos x} = \sec x}$, this term is equal to $\mathbf{\sec^2 x}$.

Simplify the second term:

$\frac{\cos^2 x}{\sin^2 x \cos^2 x} = \frac{\cancel{\cos^2 x}}{\sin^2 x \cancel{\cos^2 x}} = \frac{1}{\sin^2 x}$

Using the identity $\mathbf{\frac{1}{\sin x} = \text{cosec } x}$, this term is equal to $\mathbf{\text{cosec}^2 x}$.

So the integrand simplifies to:

$\frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} = \sec^2 x - \text{cosec}^2 x$

Now, integrate the simplified expression term by term:

$\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} \, dx = \int (\sec^2 x - \text{cosec}^2 x) \, dx$

$= \int \sec^2 x \, dx - \int \text{cosec}^2 x \, dx$

We use the standard integral formulas:

$\mathbf{\int \sec^2 x \, dx = \tan x + C_1}$

$\mathbf{\int \text{cosec}^2 x \, dx = -\cot x + C_2}$

Combining the results and adding the constant of integration $C$ ($= C_1 - C_2$):

$\int (\sec^2 x - \text{cosec}^2 x) \, dx = \tan x - (-\cot x) + C$

$= \tan x + \cot x + C$

Comparing this result with the given options:

(A) $\tan x + \cot x + C$

(B) $\tan x + \text{cosec } x + C$

(C) $-\tan x + \cot x + C$

(D) $\tan x + \sec x + C$

The obtained integral matches option (A).

The final answer is $\mathbf{\tan x + \cot x + C}$.

The correct option is (A).

To find the integral of $\int \frac{e^x (1 + x)}{\cos^2 (e^x x)}\; dx$.

Let the integral be $I = \int \frac{e^x (1 + x)}{\cos^2 (e^x x)}\; dx$.

We look for a suitable substitution. Notice the term $e^x x$ in the argument of the cosine function in the denominator, and the term $e^x (1+x)$ in the numerator.

Let's find the derivative of $e^x x$ with respect to $x$ using the product rule:

$\frac{d}{dx}(e^x x) = e^x \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(e^x)$

$= e^x \cdot 1 + x \cdot e^x$

$= e^x + xe^x$

$= e^x (1 + x)$

This derivative matches the term in the numerator of the integrand.

This suggests the substitution $\mathbf{u = e^x x}$.

Then, the differential $du$ is $\mathbf{du = e^x (1 + x) \, dx}$.

Substituting $u$ and $du$ into the integral, we get:

$I = \int \frac{du}{\cos^2 u}$

Using the identity $\mathbf{\frac{1}{\cos u} = \sec u}$, we have $\frac{1}{\cos^2 u} = \sec^2 u$.

$I = \int \sec^2 u \, du$

Now, we integrate $\sec^2 u$ with respect to $u$.

We know that $\int \sec^2 u \, du = \tan u + C$.

$I = \tan u + C$

where $C$ is the constant of integration.

Finally, substitute back $\mathbf{u = e^x x}$ to express the result in terms of $x$:

$I = \tan (e^x x) + C$

This can also be written as $\tan (xe^x) + C$.

Comparing this result with the given options:

(A) – cot (ex^x) + C

(B) tan (xe^x) + C

(C) tan (e^x) + C

(D) cot (e^x) + C

The obtained integral matches option (B).

The final answer is $\mathbf{\tan (xe^x) + C}$.

The correct option is (B).

(i)

To Find: $\int \frac{dx}{x^2 − 16}$

Solution:

The integral is of the form $\int \frac{dx}{x^2 - a^2}$.

Here, $a^2 = 16$, so $a = 4$.

We use the standard integral formula:

$\mathbf{\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + C}$

Substitute $a=4$ into the formula:

$\int \frac{dx}{x^2 - 16} = \frac{1}{2(4)} \ln \left| \frac{x-4}{x+4} \right| + C$

$= \frac{1}{8} \ln \left| \frac{x-4}{x+4} \right| + C$

Thus, the integral is:

$\mathbf{\int \frac{dx}{x^2 − 16} = \frac{1}{8} \ln \left| \frac{x-4}{x+4} \right| + C}$

(ii)

To Find: $\int \frac{dx}{\sqrt{2x − x^2}}$

Solution:

We need to complete the square for the expression under the square root in the denominator, $2x - x^2$.

$2x - x^2 = -(x^2 - 2x)$

To complete the square for $x^2 - 2x$, we add and subtract $(\frac{-2}{2})^2 = (-1)^2 = 1$.

$x^2 - 2x = (x^2 - 2x + 1) - 1 = (x-1)^2 - 1$

So, $2x - x^2 = -((x-1)^2 - 1) = 1 - (x-1)^2$.

The integral becomes:

$\int \frac{dx}{\sqrt{1 - (x-1)^2}}$

This integral is of the form $\int \frac{dx}{\sqrt{a^2 - u^2}}$, where $a=1$ and $u = x-1$.

We use the standard integral formula:

$\mathbf{\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C}$

Substitute $a=1$ and the variable $x-1$ into the formula:

$\int \frac{dx}{\sqrt{1 - (x-1)^2}} = \sin^{-1}\left(\frac{x-1}{1}\right) + C$

$= \sin^{-1}(x-1) + C$

Thus, the integral is:

$\mathbf{\int \frac{dx}{\sqrt{2x − x^2}} = \sin^{-1}(x-1) + C}$

(i)

To Find: $\int \frac{dx}{x^2 − 6x + 13}$

Solution:

We need to evaluate the integral $\int \frac{dx}{x^2 - 6x + 13}$.

The denominator is a quadratic expression. We complete the square for $x^2 - 6x + 13$.

$x^2 - 6x + 13 = (x^2 - 6x + (-3)^2) + 13 - (-3)^2$

$= (x - 3)^2 + 13 - 9$

$= (x - 3)^2 + 4$

The integral becomes:

$\int \frac{dx}{(x - 3)^2 + 4}$

This integral is of the form $\int \frac{du}{u^2 + a^2}$, where $u = x - 3$ and $a^2 = 4$, so $a = 2$.

We use the substitution $u = x - 3$, which implies $du = dx$.

The integral is $\int \frac{du}{u^2 + 2^2}$.

We use the standard integration formula:

$\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$

Substituting $u = x-3$ and $a = 2$ into the formula, we get:

$\int \frac{dx}{(x - 3)^2 + 4} = \frac{1}{2} \tan^{-1}\left(\frac{x - 3}{2}\right) + C$

Thus, the integral is:

$\mathbf{\int \frac{dx}{x^2 − 6x + 13} = \frac{1}{2} \tan^{-1}\left(\frac{x - 3}{2}\right) + C}$

(ii)

To Find: $\int \frac{dx}{3x^2 + 13x − 10}$

Solution:

We need to evaluate the integral $\int \frac{dx}{3x^2 + 13x - 10}$.

We factor out the coefficient of $x^2$ from the denominator:

$3x^2 + 13x - 10 = 3\left(x^2 + \frac{13}{3}x - \frac{10}{3}\right)$

The integral becomes: $\frac{1}{3} \int \frac{dx}{x^2 + \frac{13}{3}x - \frac{10}{3}}$.

Now, we complete the square for $x^2 + \frac{13}{3}x - \frac{10}{3}$.

We add and subtract the square of half of the coefficient of $x$: $(\frac{1}{2} \cdot \frac{13}{3})^2 = (\frac{13}{6})^2 = \frac{169}{36}$.

$x^2 + \frac{13}{3}x - \frac{10}{3} = \left(x^2 + \frac{13}{3}x + \frac{169}{36}\right) - \frac{10}{3} - \frac{169}{36}$

$= \left(x + \frac{13}{6}\right)^2 - \left(\frac{10 \cdot 12}{3 \cdot 12}\right) - \frac{169}{36}$

$= \left(x + \frac{13}{6}\right)^2 - \frac{120}{36} - \frac{169}{36}$

$= \left(x + \frac{13}{6}\right)^2 - \frac{289}{36}$

$= \left(x + \frac{13}{6}\right)^2 - \left(\frac{17}{6}\right)^2$

The integral becomes:

$\frac{1}{3} \int \frac{dx}{\left(x + \frac{13}{6}\right)^2 - \left(\frac{17}{6}\right)^2}$

This integral is of the form $\int \frac{du}{u^2 - a^2}$, where $u = x + \frac{13}{6}$ and $a = \frac{17}{6}$.

We use the substitution $u = x + \frac{13}{6}$, which implies $du = dx$.

The integral is $\frac{1}{3} \int \frac{du}{u^2 - a^2}$.

We use the standard integration formula:

$\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C'$

Substituting $u = x + \frac{13}{6}$ and $a = \frac{17}{6}$ into the formula, we get:

$\frac{1}{3} \left( \frac{1}{2\left(\frac{17}{6}\right)} \ln \left| \frac{\left(x + \frac{13}{6}\right) - \frac{17}{6}}{\left(x + \frac{13}{6}\right) + \frac{17}{6}} \right| \right) + C$

$= \frac{1}{3} \left( \frac{1}{\frac{17}{3}} \ln \left| \frac{\frac{6x + 13 - 17}{6}}{\frac{6x + 13 + 17}{6}} \right| \right) + C$

$= \frac{1}{3} \left( \frac{3}{17} \ln \left| \frac{6x - 4}{6x + 30} \right| \right) + C$

$= \frac{1}{17} \ln \left| \frac{2(3x - 2)}{2(3x + 15)} \right| + C$

$= \frac{1}{17} \ln \left| \frac{3x - 2}{3x + 15} \right| + C$

Thus, the integral is:

$\mathbf{\int \frac{dx}{3x^2 + 13x − 10} = \frac{1}{17} \ln \left| \frac{3x - 2}{3x + 15} \right| + C}$

(iii)

To Find: $\int \frac{dx}{\sqrt{5x^2 − 2x}}$

Solution:

We need to evaluate the integral $\int \frac{dx}{\sqrt{5x^2 - 2x}}$.

We factor out the coefficient of $x^2$ from the expression under the square root:

$5x^2 - 2x = 5\left(x^2 - \frac{2}{5}x\right)$

So, $\sqrt{5x^2 - 2x} = \sqrt{5\left(x^2 - \frac{2}{5}x\right)} = \sqrt{5} \sqrt{x^2 - \frac{2}{5}x}$.

The integral becomes: $\int \frac{dx}{\sqrt{5} \sqrt{x^2 - \frac{2}{5}x}} = \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{x^2 - \frac{2}{5}x}}$.

Now, we complete the square for $x^2 - \frac{2}{5}x$.

We add and subtract the square of half of the coefficient of $x$: $(\frac{1}{2} \cdot -\frac{2}{5})^2 = (-\frac{1}{5})^2 = \frac{1}{25}$.

$x^2 - \frac{2}{5}x = \left(x^2 - \frac{2}{5}x + \frac{1}{25}\right) - \frac{1}{25} = \left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2$

The integral becomes:

$\frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2}}$

This integral is of the form $\int \frac{du}{\sqrt{u^2 - a^2}}$, where $u = x - \frac{1}{5}$ and $a = \frac{1}{5}$.

We use the substitution $u = x - \frac{1}{5}$, which implies $du = dx$.

The integral is $\frac{1}{\sqrt{5}} \int \frac{du}{\sqrt{u^2 - a^2}}$.

We use the standard integration formula:

$\int \frac{du}{\sqrt{u^2 - a^2}} = \ln |u + \sqrt{u^2 - a^2}| + C'$

Substituting $u = x - \frac{1}{5}$ and $a = \frac{1}{5}$ into the formula, we get:

$\frac{1}{\sqrt{5}} \ln \left| \left(x - \frac{1}{5}\right) + \sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2} \right| + C$

Simplify the term inside the square root back to the original form (or the completed square form before taking outside the factor of 5):

$\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2 = x^2 - \frac{2}{5}x$

So, the term under the square root is $\sqrt{x^2 - \frac{2}{5}x}$.

The integral is:

$\frac{1}{\sqrt{5}} \ln \left| x - \frac{1}{5} + \sqrt{x^2 - \frac{2}{5}x} \right| + C$

Alternatively, using the formula $\int \frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \ln |2ax + b + 2\sqrt{a(ax^2+bx+c)}| + C$ for $a > 0$:

Here $a=5, b=-2, c=0$.

$\frac{1}{\sqrt{5}} \ln |2(5)x + (-2) + 2\sqrt{5(5x^2 + (-2)x + 0)}| + C$

$= \frac{1}{\sqrt{5}} \ln |10x - 2 + 2\sqrt{25x^2 - 10x}| + C$

$= \frac{1}{\sqrt{5}} \ln |2(5x - 1 + \sqrt{25x^2 - 10x})| + C$

$= \frac{1}{\sqrt{5}} (\ln 2 + \ln |5x - 1 + \sqrt{25x^2 - 10x}|) + C$

$= \frac{1}{\sqrt{5}} \ln |5x - 1 + \sqrt{25x^2 - 10x}| + C'$ (where $C' = C + \frac{\ln 2}{\sqrt{5}}$)

Both forms are equivalent.

Thus, the integral is:

$\mathbf{\int \frac{dx}{\sqrt{5x^2 − 2x}} = \frac{1}{\sqrt{5}} \ln \left| x - \frac{1}{5} + \sqrt{x^2 - \frac{2}{5}x} \right| + C}$

or equivalently

$\mathbf{\int \frac{dx}{\sqrt{5x^2 − 2x}} = \frac{1}{\sqrt{5}} \ln |5x - 1 + \sqrt{25x^2 - 10x}| + C}$

(i)

To Find: $\int \frac{x + 2}{2x^2 + 6x + 5} \;dx$

Solution:

We need to evaluate the integral $\int \frac{x + 2}{2x^2 + 6x + 5} \, dx$.

The numerator is a linear expression, and the denominator is a quadratic expression. We use the form $Px + Q = A \frac{d}{dx}(ax^2 + bx + c) + B$.

Here, $P=1, Q=2$, and the denominator is $2x^2 + 6x + 5$.

The derivative of the denominator is $\frac{d}{dx}(2x^2 + 6x + 5) = 4x + 6$.

We want to find constants $A$ and $B$ such that:

$x + 2 = A(4x + 6) + B$

$x + 2 = 4Ax + 6A + B$

Equating coefficients of $x$ and the constant terms:

From the first equation, $A = \frac{1}{4}$.

Substitute $A = \frac{1}{4}$ into the second equation:

$2 = 6\left(\frac{1}{4}\right) + B$

$2 = \frac{6}{4} + B$

$2 = \frac{3}{2} + B$

$B = 2 - \frac{3}{2} = \frac{4 - 3}{2} = \frac{1}{2}$.

So, $x + 2 = \frac{1}{4}(4x + 6) + \frac{1}{2}$.

Substitute this back into the integral:

$\int \frac{\frac{1}{4}(4x + 6) + \frac{1}{2}}{2x^2 + 6x + 5} \, dx = \int \left(\frac{\frac{1}{4}(4x + 6)}{2x^2 + 6x + 5} + \frac{\frac{1}{2}}{2x^2 + 6x + 5}\right) \, dx$

$= \frac{1}{4} \int \frac{4x + 6}{2x^2 + 6x + 5} \, dx + \frac{1}{2} \int \frac{dx}{2x^2 + 6x + 5}$

Let $I_1 = \frac{1}{4} \int \frac{4x + 6}{2x^2 + 6x + 5} \, dx$ and $I_2 = \frac{1}{2} \int \frac{dx}{2x^2 + 6x + 5}$.

Consider $I_1$. Let $u = 2x^2 + 6x + 5$. Then $du = (4x + 6) \, dx$.

$I_1 = \frac{1}{4} \int \frac{du}{u} = \frac{1}{4} \ln|u| + C_1 = \frac{1}{4} \ln|2x^2 + 6x + 5| + C_1$.

Consider $I_2$. We need to evaluate $\int \frac{dx}{2x^2 + 6x + 5}$.

Complete the square for the denominator: $2x^2 + 6x + 5 = 2(x^2 + 3x + \frac{5}{2})$.

$x^2 + 3x + \frac{5}{2} = (x^2 + 3x + (\frac{3}{2})^2) + \frac{5}{2} - (\frac{3}{2})^2$

$= (x + \frac{3}{2})^2 + \frac{5}{2} - \frac{9}{4}$

$= (x + \frac{3}{2})^2 + \frac{10 - 9}{4} = (x + \frac{3}{2})^2 + \frac{1}{4}$

So, $2x^2 + 6x + 5 = 2\left(\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2\right)$.

$I_2 = \frac{1}{2} \int \frac{dx}{2\left(\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2\right)} = \frac{1}{4} \int \frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2}$.

This is in the form $\int \frac{du}{u^2 + a^2}$ with $u = x + \frac{3}{2}$ (so $du=dx$) and $a = \frac{1}{2}$.

$I_2 = \frac{1}{4} \left( \frac{1}{\frac{1}{2}} \tan^{-1}\left(\frac{x + \frac{3}{2}}{\frac{1}{2}}\right) \right) + C_2$

$= \frac{1}{4} (2) \tan^{-1}\left(\frac{\frac{2x + 3}{2}}{\frac{1}{2}}\right) + C_2$

$= \frac{1}{2} \tan^{-1}(2x + 3) + C_2$.

Combining $I_1$ and $I_2$ and adding the constant of integration $C = C_1 + C_2$:

$\int \frac{x + 2}{2x^2 + 6x + 5} \, dx = I_1 + I_2 = \frac{1}{4} \ln|2x^2 + 6x + 5| + \frac{1}{2} \tan^{-1}(2x + 3) + C$

Thus, the integral is:

$\mathbf{\int \frac{x + 2}{2x^2 + 6x + 5} \;dx = \frac{1}{4} \ln|2x^2 + 6x + 5| + \frac{1}{2} \tan^{-1}(2x + 3) + C}$

(ii)

To Find: $\int \frac{x + 3}{\sqrt{5 − 4x − x^2}} \;dx$

Solution:

We need to evaluate the integral $\int \frac{x + 3}{\sqrt{5 - 4x - x^2}} \, dx$.

The numerator is linear, and the expression under the square root in the denominator is quadratic. We use the form $Px + Q = A \frac{d}{dx}(ax^2 + bx + c) + B$.

Here, $P=1, Q=3$, and the quadratic is $-x^2 - 4x + 5$.

The derivative of the quadratic is $\frac{d}{dx}(-x^2 - 4x + 5) = -2x - 4$.

We want to find constants $A$ and $B$ such that:

$x + 3 = A(-2x - 4) + B$

$x + 3 = -2Ax - 4A + B$

Equating coefficients of $x$ and the constant terms:

From the first equation, $A = -\frac{1}{2}$.

Substitute $A = -\frac{1}{2}$ into the second equation:

$3 = -4\left(-\frac{1}{2}\right) + B$

$3 = 2 + B$

$B = 3 - 2 = 1$.

So, $x + 3 = -\frac{1}{2}(-2x - 4) + 1$.

Substitute this back into the integral:

$\int \frac{-\frac{1}{2}(-2x - 4) + 1}{\sqrt{5 - 4x - x^2}} \, dx = \int \left(\frac{-\frac{1}{2}(-2x - 4)}{\sqrt{5 - 4x - x^2}} + \frac{1}{\sqrt{5 - 4x - x^2}}\right) \, dx$

$= -\frac{1}{2} \int \frac{-2x - 4}{\sqrt{5 - 4x - x^2}} \, dx + \int \frac{dx}{\sqrt{5 - 4x - x^2}}$

Let $I_1 = -\frac{1}{2} \int \frac{-2x - 4}{\sqrt{5 - 4x - x^2}} \, dx$ and $I_2 = \int \frac{dx}{\sqrt{5 - 4x - x^2}}$.

Consider $I_1$. Let $u = 5 - 4x - x^2$. Then $du = (-4 - 2x) \, dx = -(2x + 4) \, dx$.

$-du = (2x + 4) \, dx$. The numerator is $-2x - 4 = -(2x + 4)$, so the numerator is $-(-du) = du$.

$I_1 = -\frac{1}{2} \int \frac{du}{\sqrt{u}}$

$I_1 = -\frac{1}{2} \int u^{-1/2} \, du$

Using the power rule: $\int u^n \, du = \frac{u^{n+1}}{n+1}$.

$I_1 = -\frac{1}{2} \left( \frac{u^{-1/2 + 1}}{-1/2 + 1} \right) + C_1$

$I_1 = -\frac{1}{2} \left( \frac{u^{1/2}}{1/2} \right) + C_1$

$I_1 = -\frac{1}{2} (2\sqrt{u}) + C_1 = -\sqrt{u} + C_1$

Substitute back $u = 5 - 4x - x^2$:

$I_1 = -\sqrt{5 - 4x - x^2} + C_1$.

Consider $I_2$. We need to evaluate $\int \frac{dx}{\sqrt{5 - 4x - x^2}}$.

Complete the square for the expression under the square root: $5 - 4x - x^2 = -(x^2 + 4x - 5)$.

$x^2 + 4x - 5 = (x^2 + 4x + 2^2) - 5 - 2^2$

$= (x + 2)^2 - 5 - 4 = (x + 2)^2 - 9$

So, $5 - 4x - x^2 = -((x + 2)^2 - 9) = 9 - (x + 2)^2$.

$I_2 = \int \frac{dx}{\sqrt{9 - (x + 2)^2}}$.

This is in the form $\int \frac{du}{\sqrt{a^2 - u^2}}$ with $u = x + 2$ (so $du=dx$) and $a^2 = 9$, so $a = 3$.

Using the standard integration formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C'$:

$I_2 = \sin^{-1}\left(\frac{x + 2}{3}\right) + C_2$.

Combining $I_1$ and $I_2$ and adding the constant of integration $C = C_1 + C_2$:

$\int \frac{x + 3}{\sqrt{5 - 4x - x^2}} \, dx = I_1 + I_2 = -\sqrt{5 - 4x - x^2} + \sin^{-1}\left(\frac{x + 2}{3}\right) + C$

Thus, the integral is:

$\mathbf{\int \frac{x + 3}{\sqrt{5 − 4x − x^2}} \;dx = -\sqrt{5 - 4x - x^2} + \sin^{-1}\left(\frac{x + 2}{3}\right) + C}$

Solution:

We are asked to integrate the function $\frac{3x^2}{x^6 + 1}$.

Let the integral be $I$.

$I = \int \frac{3x^2}{x^6 + 1} dx$

We can rewrite the denominator as $(x^3)^2 + 1$. This suggests a substitution.

Let $u = x^3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$

So, $du = 3x^2 dx$.

Now substitute $u$ and $du$ into the integral $I$:

$I = \int \frac{1}{(x^3)^2 + 1} (3x^2 dx)$

$I = \int \frac{1}{u^2 + 1} du$

We know the standard integral $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$.

Here, $a=1$. So, integrating with respect to $u$, we get:

$I = \arctan(u) + C$

Substitute back $u = x^3$:

$I = \arctan(x^3) + C$

Thus, the integral of the given function is $\arctan(x^3) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{1 + 4x^2}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{1 + 4x^2}} dx$

This integral is in the form $\int \frac{1}{\sqrt{a^2 + (bx)^2}} dx$. We can use a substitution to simplify it.

Let $u = 2x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(2x) = 2$

So, $du = 2 dx$, which implies $dx = \frac{1}{2} du$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{1 + u^2}} \left(\frac{1}{2} du\right)$

$I = \frac{1}{2} \int \frac{1}{\sqrt{1^2 + u^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{a^2 + x^2}} dx$, which is $\ln|x + \sqrt{a^2 + x^2}| + C$.

Here, the variable is $u$ and $a=1$. So, integrating with respect to $u$, we get:

$I = \frac{1}{2} \ln|u + \sqrt{1^2 + u^2}| + C$

Substitute back $u = 2x$:

$I = \frac{1}{2} \ln|2x + \sqrt{1^2 + (2x)^2}| + C$

$I = \frac{1}{2} \ln|2x + \sqrt{1 + 4x^2}| + C$

Thus, the integral of the given function is $\frac{1}{2} \ln|2x + \sqrt{1 + 4x^2}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{(2 − x)^2 + 1}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{(2 − x)^2 + 1}} dx$

This integral is of the form $\int \frac{1}{\sqrt{(a - x)^2 + b^2}} dx$. We can use a substitution to simplify it.

Let $u = 2 - x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(2 - x) = -1$

So, $du = -dx$, which implies $dx = -du$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{u^2 + 1^2}} (-du)$

$I = - \int \frac{1}{\sqrt{u^2 + 1}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{x^2 + a^2}} dx$, which is $\ln|x + \sqrt{x^2 + a^2}| + C$.

Here, the variable is $u$ and $a=1$. So, integrating with respect to $u$, we get:

$I = - \left( \ln|u + \sqrt{u^2 + 1^2}| \right) + C$

$I = - \ln|u + \sqrt{u^2 + 1}| + C$

Substitute back $u = 2 - x$:

$I = - \ln|(2 - x) + \sqrt{(2 - x)^2 + 1}| + C$

Using the property of logarithms $ - \ln A = \ln(1/A) $:

$I = \ln\left|\frac{1}{(2 - x) + \sqrt{(2 - x)^2 + 1}}\right| + C$

However, the form $ - \ln|2x + \sqrt{1 + 4x^2}| $ from the previous problem suggests keeping the $ - \ln $ form is also acceptable and often standard.

Thus, the integral of the given function is $- \ln|(2 - x) + \sqrt{(2 - x)^2 + 1}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{9 − 25x^2}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{9 − 25x^2}} dx$

We can rewrite the denominator as $\sqrt{3^2 - (5x)^2}$. This integral is in the form $\int \frac{1}{\sqrt{a^2 - (bx)^2}} dx$. We can use a substitution.

Let $u = 5x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(5x) = 5$

So, $du = 5 dx$, which implies $dx = \frac{1}{5} du$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{3^2 - u^2}} \left(\frac{1}{5} du\right)$

$I = \frac{1}{5} \int \frac{1}{\sqrt{3^2 - u^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{a^2 - x^2}} dx$, which is $\arcsin\left(\frac{x}{a}\right) + C$.

Here, the variable is $u$ and $a=3$. So, integrating with respect to $u$, we get:

$I = \frac{1}{5} \arcsin\left(\frac{u}{3}\right) + C$

Substitute back $u = 5x$:

$I = \frac{1}{5} \arcsin\left(\frac{5x}{3}\right) + C$

Thus, the integral of the given function is $\frac{1}{5} \arcsin\left(\frac{5x}{3}\right) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{3x}{1 + 2x^4}$.

Let the integral be $I$.

$I = \int \frac{3x}{1 + 2x^4} dx$

We can rewrite the denominator as $1 + 2(x^2)^2$. This suggests a substitution involving $x^2$.

Let $u = x^2$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x$

So, $du = 2x dx$.

We have $3x dx$ in the numerator. We can write $3x dx = \frac{3}{2} (2x dx) = \frac{3}{2} du$.

Now substitute $u$ and $du$ into the integral $I$:

$I = \int \frac{1}{1 + 2(x^2)^2} (3x dx)$

$I = \int \frac{1}{1 + 2u^2} \left(\frac{3}{2} du\right)$

$I = \frac{3}{2} \int \frac{1}{1 + 2u^2} du$

We can write $1 + 2u^2$ as $1^2 + (\sqrt{2}u)^2$. The integral is in the form $\int \frac{1}{a^2 + (bu)^2} du$.

Using the standard integral formula $\int \frac{1}{a^2 + (bx)^2} dx = \frac{1}{ab} \arctan\left(\frac{bx}{a}\right) + C$, with $a=1$, $b=\sqrt{2}$ and variable $u$ instead of $x$:

$I = \frac{3}{2} \left( \frac{1}{1 \cdot \sqrt{2}} \arctan\left(\frac{\sqrt{2}u}{1}\right) \right) + C'$

$I = \frac{3}{2\sqrt{2}} \arctan(\sqrt{2}u) + C'$

Substitute back $u = x^2$:

$I = \frac{3}{2\sqrt{2}} \arctan(\sqrt{2}x^2) + C'$

The term $\frac{3}{2\sqrt{2}}$ can be rationalized as $\frac{3\sqrt{2}}{4}$.

$I = \frac{3\sqrt{2}}{4} \arctan(\sqrt{2}x^2) + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\frac{3\sqrt{2}}{4} \arctan(\sqrt{2}x^2) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{x^2}{1 − x^6}$.

Let the integral be $I$.

$I = \int \frac{x^2}{1 − x^6} dx$

We can rewrite the denominator as $1 - (x^3)^2$. This suggests a substitution involving $x^3$.

Let $u = x^3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$

So, $du = 3x^2 dx$.

We have $x^2 dx$ in the numerator. We can write $x^2 dx = \frac{1}{3} (3x^2 dx) = \frac{1}{3} du$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{1 - (x^3)^2} (x^2 dx)$

$I = \int \frac{1}{1 - u^2} \left(\frac{1}{3} du\right)$

$I = \frac{1}{3} \int \frac{1}{1^2 - u^2} du$

We use the standard integral formula for the form $\int \frac{1}{a^2 - x^2} dx$, which is $\frac{1}{2a} \ln\left|\frac{a+x}{a-x}\right| + C$.

Here, the variable is $u$ and $a=1$. So, integrating with respect to $u$, we get:

$I = \frac{1}{3} \left( \frac{1}{2 \cdot 1} \ln\left|\frac{1+u}{1-u}\right| \right) + C'$

$I = \frac{1}{6} \ln\left|\frac{1+u}{1-u}\right| + C'$

Substitute back $u = x^3$:

$I = \frac{1}{6} \ln\left|\frac{1+x^3}{1-x^3}\right| + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\frac{1}{6} \ln\left|\frac{1+x^3}{1-x^3}\right| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{x − 1}{\sqrt{x^2 − 1}}$.

Let the integral be $I$.

$I = \int \frac{x − 1}{\sqrt{x^2 − 1}} dx$

We can split the integral into two parts based on the numerator:

$I = \int \frac{x}{\sqrt{x^2 − 1}} dx - \int \frac{1}{\sqrt{x^2 − 1}} dx$

Let's evaluate the first integral: $I_1 = \int \frac{x}{\sqrt{x^2 − 1}} dx$.

Let $u = x^2 - 1$.

Then $du = 2x dx$, so $x dx = \frac{1}{2} du$.

$I_1 = \int \frac{1}{\sqrt{u}} \left(\frac{1}{2} du\right) = \frac{1}{2} \int u^{-1/2} du$

Integrating $u^{-1/2}$ gives $\frac{u^{-1/2 + 1}}{-1/2 + 1} = \frac{u^{1/2}}{1/2} = 2\sqrt{u}$.

$I_1 = \frac{1}{2} (2\sqrt{u}) + C_1 = \sqrt{u} + C_1$

Substituting back $u = x^2 - 1$:

$I_1 = \sqrt{x^2 - 1} + C_1$

Now let's evaluate the second integral: $I_2 = \int \frac{1}{\sqrt{x^2 − 1}} dx$.

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln|x + \sqrt{x^2 - a^2}| + C$.

Here, $a=1$.

$I_2 = \ln|x + \sqrt{x^2 - 1^2}| + C_2 = \ln|x + \sqrt{x^2 - 1}| + C_2$

Combining the two integrals $I = I_1 - I_2$:

$I = (\sqrt{x^2 - 1} + C_1) - (\ln|x + \sqrt{x^2 - 1}| + C_2)$

$I = \sqrt{x^2 - 1} - \ln|x + \sqrt{x^2 - 1}| + (C_1 - C_2)$

Let $C = C_1 - C_2$ be the constant of integration.

$I = \sqrt{x^2 - 1} - \ln|x + \sqrt{x^2 - 1}| + C$

Thus, the integral of the given function is $\sqrt{x^2 - 1} - \ln|x + \sqrt{x^2 - 1}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{x^2}{\sqrt{x^6 + a^6}}$.

Let the integral be $I$.

$I = \int \frac{x^2}{\sqrt{x^6 + a^6}} dx$

We can rewrite the denominator as $\sqrt{(x^3)^2 + (a^3)^2}$. This suggests a substitution involving $x^3$.

Let $u = x^3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$

So, $du = 3x^2 dx$.

We have $x^2 dx$ in the numerator. We can write $x^2 dx = \frac{1}{3} (3x^2 dx) = \frac{1}{3} du$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{(x^3)^2 + (a^3)^2}} (x^2 dx)$

$I = \int \frac{1}{\sqrt{u^2 + (a^3)^2}} \left(\frac{1}{3} du\right)$

$I = \frac{1}{3} \int \frac{1}{\sqrt{u^2 + (a^3)^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{x^2 + k^2}} dx$, which is $\ln|x + \sqrt{x^2 + k^2}| + C$.

Here, the variable is $u$ and $k = a^3$. So, integrating with respect to $u$, we get:

$I = \frac{1}{3} \ln|u + \sqrt{u^2 + (a^3)^2}| + C'$

Substitute back $u = x^3$:

$I = \frac{1}{3} \ln|x^3 + \sqrt{(x^3)^2 + (a^3)^2}| + C'$

$I = \frac{1}{3} \ln|x^3 + \sqrt{x^6 + a^6}| + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\frac{1}{3} \ln|x^3 + \sqrt{x^6 + a^6}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}$.

Let the integral be $I$.

$I = \int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} dx$

This integral suggests a substitution involving $\tan x$ because the numerator $\sec^2 x$ is the derivative of $\tan x$.

Let $u = \tan x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

So, $du = \sec^2 x dx$.

Now substitute $u$ and $du$ into the integral $I$:

$I = \int \frac{1}{\sqrt{(\tan x)^2 + 4}} (\sec^2 x dx)$

$I = \int \frac{1}{\sqrt{u^2 + 4}} du$

$I = \int \frac{1}{\sqrt{u^2 + 2^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{x^2 + a^2}} dx$, which is $\ln|x + \sqrt{x^2 + a^2}| + C$.

Here, the variable is $u$ and $a=2$. So, integrating with respect to $u$, we get:

$I = \ln|u + \sqrt{u^2 + 2^2}| + C'$

$I = \ln|u + \sqrt{u^2 + 4}| + C'$

Substitute back $u = \tan x$:

$I = \ln|\tan x + \sqrt{\tan^2 x + 4}| + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\ln|\tan x + \sqrt{\tan^2 x + 4}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{x^2 + 2x + 2}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx$

The expression under the square root in the denominator is a quadratic. We complete the square for $x^2 + 2x + 2$.

$x^2 + 2x + 2 = (x^2 + 2x + 1) - 1 + 2$

$x^2 + 2x + 2 = (x+1)^2 + 1$

Substitute the completed square form into the integral:

$I = \int \frac{1}{\sqrt{(x+1)^2 + 1}} dx$

This integral is in the form $\int \frac{1}{\sqrt{u^2 + a^2}} du$. We can use a substitution.

Let $u = x+1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x+1) = 1$

So, $du = dx$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{u^2 + 1^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{x^2 + a^2}} dx$, which is $\ln|x + \sqrt{x^2 + a^2}| + C$.

Here, the variable is $u$ and $a=1$. So, integrating with respect to $u$, we get:

$I = \ln|u + \sqrt{u^2 + 1^2}| + C'$

$I = \ln|u + \sqrt{u^2 + 1}| + C'$

Substitute back $u = x+1$:

$I = \ln|(x+1) + \sqrt{(x+1)^2 + 1}| + C'$

Simplify the expression under the square root:

$(x+1)^2 + 1 = x^2 + 2x + 1 + 1 = x^2 + 2x + 2$

$I = \ln|x+1 + \sqrt{x^2 + 2x + 2}| + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\ln|x+1 + \sqrt{x^2 + 2x + 2}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{9x^2 + 6x + 5}$.

Let the integral be $I$.

$I = \int \frac{1}{9x^2 + 6x + 5} dx$

The denominator is a quadratic expression. We complete the square for $9x^2 + 6x + 5$.

Factor out 9 from the quadratic term:

$9x^2 + 6x + 5 = 9\left(x^2 + \frac{6}{9}x + \frac{5}{9}\right)$

$9x^2 + 6x + 5 = 9\left(x^2 + \frac{2}{3}x + \frac{5}{9}\right)$

Complete the square for the expression inside the parentheses $x^2 + \frac{2}{3}x$: Add and subtract $\left(\frac{1}{2} \cdot \frac{2}{3}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.

$x^2 + \frac{2}{3}x + \frac{5}{9} = \left(x^2 + \frac{2}{3}x + \frac{1}{9}\right) - \frac{1}{9} + \frac{5}{9}$

$x^2 + \frac{2}{3}x + \frac{5}{9} = \left(x + \frac{1}{3}\right)^2 + \frac{4}{9}$

So, the denominator is $9\left(\left(x + \frac{1}{3}\right)^2 + \frac{4}{9}\right)$.

The integral becomes:

$I = \int \frac{1}{9\left(\left(x + \frac{1}{3}\right)^2 + \frac{4}{9}\right)} dx$

$I = \frac{1}{9} \int \frac{1}{\left(x + \frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2} dx$

This integral is in the form $\int \frac{1}{u^2 + a^2} du$. We use a substitution.

Let $u = x + \frac{1}{3}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}\left(x + \frac{1}{3}\right) = 1$

So, $du = dx$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \frac{1}{9} \int \frac{1}{u^2 + \left(\frac{2}{3}\right)^2} du$

We use the standard integral formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$.

Here, the variable is $u$ and $a = \frac{2}{3}$. So, integrating with respect to $u$, we get:

$I = \frac{1}{9} \left( \frac{1}{\frac{2}{3}} \arctan\left(\frac{u}{\frac{2}{3}}\right) \right) + C'$

$I = \frac{1}{9} \left( \frac{3}{2} \arctan\left(\frac{3u}{2}\right) \right) + C'$

$I = \frac{3}{18} \arctan\left(\frac{3u}{2}\right) + C'$

$I = \frac{1}{6} \arctan\left(\frac{3u}{2}\right) + C'$

Substitute back $u = x + \frac{1}{3}$:

$I = \frac{1}{6} \arctan\left(\frac{3\left(x + \frac{1}{3}\right)}{2}\right) + C'$

$I = \frac{1}{6} \arctan\left(\frac{3x + 1}{2}\right) + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\frac{1}{6} \arctan\left(\frac{3x + 1}{2}\right) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{7 − 6x − x^2}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{7 − 6x − x^2}} dx$

The expression under the square root in the denominator is a quadratic. We complete the square for $7 − 6x − x^2$.

First, rewrite the quadratic term by factoring out $-1$ from the $x$ terms:

$7 − 6x − x^2 = 7 - (x^2 + 6x)$

Complete the square for the expression inside the parentheses $x^2 + 6x$. Add and subtract $\left(\frac{1}{2} \cdot 6\right)^2 = 3^2 = 9$.

$x^2 + 6x = (x^2 + 6x + 9) - 9 = (x+3)^2 - 9$

Now substitute this back into the original expression:

$7 - (x^2 + 6x) = 7 - ((x+3)^2 - 9)$

$= 7 - (x+3)^2 + 9$

$= 16 - (x+3)^2$

So, the expression under the square root is $16 - (x+3)^2$. The integral becomes:

$I = \int \frac{1}{\sqrt{16 - (x+3)^2}} dx$

$I = \int \frac{1}{\sqrt{4^2 - (x+3)^2}} dx$

This integral is in the form $\int \frac{1}{\sqrt{a^2 - u^2}} du$. We use a substitution.

Let $u = x+3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x+3) = 1$

So, $du = dx$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{4^2 - u^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{a^2 - x^2}} dx$, which is $\arcsin\left(\frac{x}{a}\right) + C$.

Here, the variable is $u$ and $a=4$. So, integrating with respect to $u$, we get:

$I = \arcsin\left(\frac{u}{4}\right) + C'$

Substitute back $u = x+3$:

$I = \arcsin\left(\frac{x+3}{4}\right) + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\arcsin\left(\frac{x+3}{4}\right) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{(x − 1) (x − 2)}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{(x − 1) (x − 2)}} dx$

First, we simplify the expression under the square root in the denominator by expanding the product:

$(x - 1)(x - 2) = x^2 - 2x - x + 2 = x^2 - 3x + 2$

Now, we complete the square for the quadratic expression $x^2 - 3x + 2$.

The coefficient of $x$ is $-3$. Half of this is $-\frac{3}{2}$. The square of half the coefficient is $\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$.

We add and subtract $\frac{9}{4}$ inside the expression:

$x^2 - 3x + 2 = \left(x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} + 2$

$x^2 - 3x + 2 = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} + \frac{8}{4}$

$x^2 - 3x + 2 = \left(x - \frac{3}{2}\right)^2 - \frac{1}{4}$

$x^2 - 3x + 2 = \left(x - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2$

So, the integral becomes:

$I = \int \frac{1}{\sqrt{\left(x - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} dx$

This integral is in the form $\int \frac{1}{\sqrt{u^2 - a^2}} du$. We can use a substitution.

Let $u = x - \frac{3}{2}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}\left(x - \frac{3}{2}\right) = 1$

So, $du = dx$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{u^2 - \left(\frac{1}{2}\right)^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx$, which is $\ln|x + \sqrt{x^2 - a^2}| + C$.

Here, the variable is $u$ and $a=\frac{1}{2}$. So, integrating with respect to $u$, we get:

$I = \ln\left|u + \sqrt{u^2 - \left(\frac{1}{2}\right)^2}\right| + C'$

$I = \ln\left|u + \sqrt{u^2 - \frac{1}{4}}\right| + C'$

Substitute back $u = x - \frac{3}{2}$:

$I = \ln\left|\left(x - \frac{3}{2}\right) + \sqrt{\left(x - \frac{3}{2}\right)^2 - \frac{1}{4}}\right| + C'$

We know that $\left(x - \frac{3}{2}\right)^2 - \frac{1}{4} = x^2 - 3x + 2 = (x-1)(x-2)$.

$I = \ln\left|x - \frac{3}{2} + \sqrt{x^2 - 3x + 2}\right| + C$

or

$I = \ln\left|x - \frac{3}{2} + \sqrt{(x-1)(x-2)}\right| + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\ln\left|x - \frac{3}{2} + \sqrt{x^2 - 3x + 2}\right| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{8 + 3x − x^2}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{8 + 3x − x^2}} dx$

The expression under the square root in the denominator is a quadratic. We complete the square for $8 + 3x − x^2$.

First, rewrite the quadratic term by factoring out $-1$ from the $x$ terms:

$8 + 3x − x^2 = 8 - (x^2 - 3x)$

Complete the square for the expression inside the parentheses $x^2 - 3x$. Add and subtract $\left(\frac{1}{2} \cdot (-3)\right)^2 = \left(-\frac{3}{2}\right)^2 = \frac{9}{4}$.

$x^2 - 3x = \left(x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$

Now substitute this back into the original expression:

$8 - (x^2 - 3x) = 8 - \left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right)$

$= 8 - \left(x - \frac{3}{2}\right)^2 + \frac{9}{4}$

$= \frac{32}{4} + \frac{9}{4} - \left(x - \frac{3}{2}\right)^2$

$= \frac{41}{4} - \left(x - \frac{3}{2}\right)^2$

$= \left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$

So, the expression under the square root is $\left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$. The integral becomes:

$I = \int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2}} dx$

This integral is in the form $\int \frac{1}{\sqrt{a^2 - u^2}} du$. We can use a substitution.

Let $u = x - \frac{3}{2}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}\left(x - \frac{3}{2}\right) = 1$

So, $du = dx$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2 - u^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{a^2 - x^2}} dx$, which is $\arcsin\left(\frac{x}{a}\right) + C$.

Here, the variable is $u$ and $a=\frac{\sqrt{41}}{2}$. So, integrating with respect to $u$, we get:

$I = \arcsin\left(\frac{u}{\frac{\sqrt{41}}{2}}\right) + C'$

$I = \arcsin\left(\frac{2u}{\sqrt{41}}\right) + C'$

Substitute back $u = x - \frac{3}{2}$:

$I = \arcsin\left(\frac{2\left(x - \frac{3}{2}\right)}{\sqrt{41}}\right) + C'$

$I = \arcsin\left(\frac{2x - 3}{\sqrt{41}}\right) + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\arcsin\left(\frac{2x - 3}{\sqrt{41}}\right) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{1}{\sqrt{(x − a) (x − b)}}$.

Let the integral be $I$.

$I = \int \frac{1}{\sqrt{(x − a) (x − b)}} dx$

First, we expand the expression under the square root in the denominator:

$(x - a)(x - b) = x^2 - bx - ax + ab = x^2 - (a+b)x + ab$

Now, we complete the square for the quadratic expression $x^2 - (a+b)x + ab$.

The coefficient of $x$ is $-(a+b)$. Half of this is $-\frac{a+b}{2}$. The square of half the coefficient is $\left(-\frac{a+b}{2}\right)^2 = \left(\frac{a+b}{2}\right)^2$.

We add and subtract $\left(\frac{a+b}{2}\right)^2$ inside the expression:

$x^2 - (a+b)x + ab = \left(x^2 - (a+b)x + \left(\frac{a+b}{2}\right)^2\right) - \left(\frac{a+b}{2}\right)^2 + ab$

$x^2 - (a+b)x + ab = \left(x - \frac{a+b}{2}\right)^2 - \frac{(a+b)^2 - 4ab}{4}$

$x^2 - (a+b)x + ab = \left(x - \frac{a+b}{2}\right)^2 - \frac{a^2 + 2ab + b^2 - 4ab}{4}$

$x^2 - (a+b)x + ab = \left(x - \frac{a+b}{2}\right)^2 - \frac{a^2 - 2ab + b^2}{4}$

$x^2 - (a+b)x + ab = \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2$

So, the integral becomes:

$I = \int \frac{1}{\sqrt{\left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2}} dx$

This integral is in the form $\int \frac{1}{\sqrt{u^2 - k^2}} du$. We use a substitution.

Let $u = x - \frac{a+b}{2}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}\left(x - \frac{a+b}{2}\right) = 1$

So, $du = dx$.

Now substitute $u$ and $dx$ into the integral $I$:

$I = \int \frac{1}{\sqrt{u^2 - \left(\frac{a-b}{2}\right)^2}} du$

We use the standard integral formula for the form $\int \frac{1}{\sqrt{x^2 - k^2}} dx$, which is $\ln|x + \sqrt{x^2 - k^2}| + C$.

Here, the variable is $u$ and $k=\frac{a-b}{2}$. So, integrating with respect to $u$, we get:

$I = \ln\left|u + \sqrt{u^2 - \left(\frac{a-b}{2}\right)^2}\right| + C'$

Substitute back $u = x - \frac{a+b}{2}$ and the completed square form:

$I = \ln\left|\left(x - \frac{a+b}{2}\right) + \sqrt{(x-a)(x-b)}\right| + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $\ln\left|x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)}\right| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{4x + 1}{\sqrt{2x^2 + x - 3}}$.

Let the integral be $I$.

$I = \int \frac{4x + 1}{\sqrt{2x^2 + x - 3}} dx$

We observe that the numerator $4x + 1$ is the derivative of the expression inside the square root in the denominator, $2x^2 + x - 3$.

Let $u = 2x^2 + x - 3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(2x^2 + x - 3) = 4x + 1$

So, $du = (4x + 1) dx$.

Now substitute $u$ and $du$ into the integral $I$:

$I = \int \frac{1}{\sqrt{2x^2 + x - 3}} (4x + 1) dx$

$I = \int \frac{1}{\sqrt{u}} du$

$I = \int u^{-1/2} du$

Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), we have:

$I = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C'$

$I = \frac{u^{1/2}}{1/2} + C'$

$I = 2\sqrt{u} + C'$

Substitute back $u = 2x^2 + x - 3$:

$I = 2\sqrt{2x^2 + x - 3} + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Thus, the integral of the given function is $2\sqrt{2x^2 + x - 3} + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{x + 2}{\sqrt{x^2 − 1}}$.

Let the integral be $I$.

$I = \int \frac{x + 2}{\sqrt{x^2 − 1}} dx$

We can split the integral into two parts based on the numerator:

$I = \int \frac{x}{\sqrt{x^2 − 1}} dx + \int \frac{2}{\sqrt{x^2 − 1}} dx$

Let's evaluate the first integral: $I_1 = \int \frac{x}{\sqrt{x^2 − 1}} dx$.

Let $u = x^2 - 1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^2 - 1) = 2x$

So, $du = 2x dx$, which implies $x dx = \frac{1}{2} du$.

Now substitute $u$ and $dx$ into $I_1$:

$I_1 = \int \frac{1}{\sqrt{u}} \left(\frac{1}{2} du\right)$

$I_1 = \frac{1}{2} \int u^{-1/2} du$

Integrating $u^{-1/2}$ gives $\frac{u^{-1/2 + 1}}{-1/2 + 1} = \frac{u^{1/2}}{1/2} = 2\sqrt{u}$.

$I_1 = \frac{1}{2} (2\sqrt{u}) + C_1 = \sqrt{u} + C_1$

Substituting back $u = x^2 - 1$:

$I_1 = \sqrt{x^2 - 1} + C_1$

Now let's evaluate the second integral: $I_2 = \int \frac{2}{\sqrt{x^2 − 1}} dx$.

$I_2 = 2 \int \frac{1}{\sqrt{x^2 − 1}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln|x + \sqrt{x^2 - a^2}| + C$.

Here, $a=1$.

$I_2 = 2 \left( \ln|x + \sqrt{x^2 - 1^2}| \right) + C_2$

$I_2 = 2 \ln|x + \sqrt{x^2 - 1}| + C_2$

Combining the two integrals $I = I_1 + I_2$:

$I = (\sqrt{x^2 - 1} + C_1) + (2 \ln|x + \sqrt{x^2 - 1}| + C_2)$

$I = \sqrt{x^2 - 1} + 2 \ln|x + \sqrt{x^2 - 1}| + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

$I = \sqrt{x^2 - 1} + 2 \ln|x + \sqrt{x^2 - 1}| + C$

Thus, the integral of the given function is $\sqrt{x^2 - 1} + 2 \ln|x + \sqrt{x^2 - 1}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{5x − 2}{1 + 2x + 3x^2}$.

Let the integral be $I$.

$I = \int \frac{5x − 2}{3x^2 + 2x + 1} dx$

For an integral of the form $\int \frac{Px+Q}{ax^2+bx+c} dx$, we express the numerator as $A \cdot \frac{d}{dx}(ax^2+bx+c) + B$.

The derivative of the denominator $3x^2 + 2x + 1$ is $6x + 2$.

We want to find constants $A$ and $B$ such that $5x - 2 = A(6x + 2) + B$.

$5x - 2 = 6Ax + 2A + B$

Comparing coefficients of $x$ on both sides:

$6A = 5 \implies A = \frac{5}{6}$

Comparing constant terms:

$2A + B = -2$

Substitute $A = \frac{5}{6}$:

$2\left(\frac{5}{6}\right) + B = -2$

$\frac{10}{6} + B = -2$

$\frac{5}{3} + B = -2$

$B = -2 - \frac{5}{3} = -\frac{6}{3} - \frac{5}{3} = -\frac{11}{3}$

So, the numerator $5x - 2$ can be written as $\frac{5}{6}(6x + 2) - \frac{11}{3}$.

Substitute this back into the integral:

$I = \int \frac{\frac{5}{6}(6x + 2) - \frac{11}{3}}{3x^2 + 2x + 1} dx$

$I = \int \left(\frac{5}{6} \cdot \frac{6x + 2}{3x^2 + 2x + 1} - \frac{11}{3} \cdot \frac{1}{3x^2 + 2x + 1}\right) dx$

$I = \frac{5}{6} \int \frac{6x + 2}{3x^2 + 2x + 1} dx - \frac{11}{3} \int \frac{1}{3x^2 + 2x + 1} dx$

Let $I_1 = \int \frac{6x + 2}{3x^2 + 2x + 1} dx$ and $I_2 = \int \frac{1}{3x^2 + 2x + 1} dx$.

For $I_1$, let $u = 3x^2 + 2x + 1$. Then $du = (6x + 2) dx$.

$I_1 = \int \frac{1}{u} du = \ln|u| + C_1 = \ln|3x^2 + 2x + 1| + C_1$.

Since the discriminant of $3x^2 + 2x + 1$ is $2^2 - 4(3)(1) = 4 - 12 = -8 < 0$ and the leading coefficient is positive, the quadratic $3x^2 + 2x + 1$ is always positive. So, $|3x^2 + 2x + 1| = 3x^2 + 2x + 1$.

$I_1 = \ln(3x^2 + 2x + 1) + C_1$.

For $I_2$, we complete the square for the denominator $3x^2 + 2x + 1$:

$3x^2 + 2x + 1 = 3\left(x^2 + \frac{2}{3}x + \frac{1}{3}\right)$

$= 3\left(x^2 + \frac{2}{3}x + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \frac{1}{3}\right)$

$= 3\left(\left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \frac{3}{9}\right)$

$= 3\left(\left(x + \frac{1}{3}\right)^2 + \frac{2}{9}\right)$

$= 3\left(\left(x + \frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2\right)$

So, $I_2 = \int \frac{1}{3\left(\left(x + \frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2\right)} dx = \frac{1}{3} \int \frac{1}{\left(x + \frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2} dx$.

Let $v = x + \frac{1}{3}$, so $dv = dx$. The integral becomes $\frac{1}{3} \int \frac{1}{v^2 + \left(\frac{\sqrt{2}}{3}\right)^2} dv$.

Using the standard integral formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$ with $a = \frac{\sqrt{2}}{3}$:

$I_2 = \frac{1}{3} \cdot \frac{1}{\frac{\sqrt{2}}{3}} \arctan\left(\frac{v}{\frac{\sqrt{2}}{3}}\right) + C_2'$

$I_2 = \frac{1}{3} \cdot \frac{3}{\sqrt{2}} \arctan\left(\frac{3v}{\sqrt{2}}\right) + C_2'$

$I_2 = \frac{1}{\sqrt{2}} \arctan\left(\frac{3\left(x + \frac{1}{3}\right)}{\sqrt{2}}\right) + C_2'$

$I_2 = \frac{1}{\sqrt{2}} \arctan\left(\frac{3x + 1}{\sqrt{2}}\right) + C_2'$

Now combine $I_1$ and $I_2$ to find the total integral $I = \frac{5}{6} I_1 - \frac{11}{3} I_2$:

$I = \frac{5}{6} \left(\ln(3x^2 + 2x + 1) + C_1\right) - \frac{11}{3} \left(\frac{1}{\sqrt{2}} \arctan\left(\frac{3x + 1}{\sqrt{2}}\right) + C_2'\right)$

$I = \frac{5}{6} \ln(3x^2 + 2x + 1) - \frac{11}{3\sqrt{2}} \arctan\left(\frac{3x + 1}{\sqrt{2}}\right) + \left(\frac{5}{6}C_1 - \frac{11}{3}C_2'\right)$

Let $C = \frac{5}{6}C_1 - \frac{11}{3}C_2'$ be the constant of integration.

$I = \frac{5}{6} \ln(3x^2 + 2x + 1) - \frac{11\sqrt{2}}{6} \arctan\left(\frac{3x + 1}{\sqrt{2}}\right) + C$

Thus, the integral of the given function is $\frac{5}{6} \ln(3x^2 + 2x + 1) - \frac{11\sqrt{2}}{6} \arctan\left(\frac{3x + 1}{\sqrt{2}}\right) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{6x + 7}{\sqrt{(x − 5)(x − 4)}}$.

Let the integral be $I$.

$I = \int \frac{6x + 7}{\sqrt{(x − 5)(x − 4)}} dx$

First, expand the expression under the square root in the denominator:

$(x − 5)(x − 4) = x^2 - 4x - 5x + 20 = x^2 - 9x + 20$

The integral is now $I = \int \frac{6x + 7}{\sqrt{x^2 - 9x + 20}} dx$.

For an integral of the form $\int \frac{Px+Q}{\sqrt{ax^2+bx+c}} dx$, we express the numerator as $A \cdot \frac{d}{dx}(ax^2+bx+c) + B$.

The derivative of the denominator's quadratic part $x^2 - 9x + 20$ is $2x - 9$.

We want to find constants $A$ and $B$ such that $6x + 7 = A(2x - 9) + B$.

Expanding the right side: $6x + 7 = 2Ax - 9A + B$

Equating coefficients of $x$:

$2A = 6 \implies A = 3$

Equating constant terms:

$-9A + B = 7$

Substitute the value of $A$:

$-9(3) + B = 7$

$-27 + B = 7$

$B = 7 + 27 = 34$

So, the numerator $6x + 7$ can be written as $3(2x - 9) + 34$.

Substitute this back into the integral:

$I = \int \frac{3(2x - 9) + 34}{\sqrt{x^2 - 9x + 20}} dx$

Split the integral into two parts:

$I = \int \frac{3(2x - 9)}{\sqrt{x^2 - 9x + 20}} dx + \int \frac{34}{\sqrt{x^2 - 9x + 20}} dx$

$I = 3 \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} dx + 34 \int \frac{1}{\sqrt{x^2 - 9x + 20}} dx$

Let $I_1 = \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} dx$ and $I_2 = \int \frac{1}{\sqrt{x^2 - 9x + 20}} dx$.

Evaluate $I_1$:

In $I_1$, let $u = x^2 - 9x + 20$.

Then $du = (2x - 9) dx$.

$I_1 = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du$

$I_1 = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_1 = \frac{u^{1/2}}{1/2} + C_1 = 2\sqrt{u} + C_1$

Substitute back $u = x^2 - 9x + 20$:

$I_1 = 2\sqrt{x^2 - 9x + 20} + C_1$

Evaluate $I_2$:

$I_2 = \int \frac{1}{\sqrt{x^2 - 9x + 20}} dx$. We complete the square for the denominator $x^2 - 9x + 20$.

$x^2 - 9x + 20 = x^2 - 9x + \left(\frac{9}{2}\right)^2 - \left(\frac{9}{2}\right)^2 + 20$

$x^2 - 9x + 20 = \left(x - \frac{9}{2}\right)^2 - \frac{81}{4} + \frac{80}{4}$

$x^2 - 9x + 20 = \left(x - \frac{9}{2}\right)^2 - \frac{1}{4}$

$x^2 - 9x + 20 = \left(x - \frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2$

So, $I_2 = \int \frac{1}{\sqrt{\left(x - \frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} dx$.

This is in the standard form $\int \frac{1}{\sqrt{u^2 - a^2}} du = \ln|u + \sqrt{u^2 - a^2}| + C$.

Here, $u = x - \frac{9}{2}$ and $a = \frac{1}{2}$.

$I_2 = \ln\left|\left(x - \frac{9}{2}\right) + \sqrt{\left(x - \frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2}\right| + C_2$

Substitute back the completed square form $x^2 - 9x + 20$:

$I_2 = \ln\left|x - \frac{9}{2} + \sqrt{x^2 - 9x + 20}\right| + C_2$

Combine $I_1$ and $I_2$ to find the total integral $I = 3 I_1 + 34 I_2$:

$I = 3 \left(2\sqrt{x^2 - 9x + 20} + C_1\right) + 34 \left(\ln\left|x - \frac{9}{2} + \sqrt{x^2 - 9x + 20}\right| + C_2\right)$

$I = 6\sqrt{x^2 - 9x + 20} + 6C_1 + 34 \ln\left|x - \frac{9}{2} + \sqrt{x^2 - 9x + 20}\right| + 34C_2$

Let $C = 6C_1 + 34C_2$ be the constant of integration.

$I = 6\sqrt{x^2 - 9x + 20} + 34 \ln\left|x - \frac{9}{2} + \sqrt{x^2 - 9x + 20}\right| + C$

The term $\sqrt{x^2 - 9x + 20}$ can also be written as $\sqrt{(x-5)(x-4)}$.

$I = 6\sqrt{(x-5)(x-4)} + 34 \ln\left|x - \frac{9}{2} + \sqrt{(x-5)(x-4)}\right| + C$

Thus, the integral of the given function is $6\sqrt{(x-5)(x-4)} + 34 \ln\left|x - \frac{9}{2} + \sqrt{(x-5)(x-4)}\right| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{x + 2}{\sqrt{4x − x^2}}$.

Let the integral be $I$.

$I = \int \frac{x + 2}{\sqrt{4x − x^2}} dx$

We express the numerator $x+2$ as $A \cdot \frac{d}{dx}(4x-x^2) + B$.

The derivative of $4x - x^2$ is $4 - 2x$.

So, we need to find constants $A$ and $B$ such that:

$x + 2 = A(4 - 2x) + B$

$x + 2 = 4A - 2Ax + B$

Comparing the coefficients of $x$:

$1 = -2A \implies A = -\frac{1}{2}$

Comparing the constant terms:

$2 = 4A + B$

Substitute $A = -\frac{1}{2}$ into this equation:

$2 = 4\left(-\frac{1}{2}\right) + B$

$2 = -2 + B$

$B = 2 + 2 = 4$

So, the numerator $x + 2$ can be written as $-\frac{1}{2}(4 - 2x) + 4$.

Substitute this back into the integral:

$I = \int \frac{-\frac{1}{2}(4 - 2x) + 4}{\sqrt{4x - x^2}} dx$

Split the integral into two parts:

$I = \int \frac{-\frac{1}{2}(4 - 2x)}{\sqrt{4x - x^2}} dx + \int \frac{4}{\sqrt{4x - x^2}} dx$

$I = -\frac{1}{2} \int \frac{4 - 2x}{\sqrt{4x - x^2}} dx + 4 \int \frac{1}{\sqrt{4x - x^2}} dx$

Let $I_1 = \int \frac{4 - 2x}{\sqrt{4x - x^2}} dx$ and $I_2 = \int \frac{1}{\sqrt{4x - x^2}} dx$.

Evaluate $I_1$:

In $I_1$, let $u = 4x - x^2$.

Then $du = (4 - 2x) dx$.

$I_1 = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du$

Using the power rule for integration:

$I_1 = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_1 = \frac{u^{1/2}}{1/2} + C_1 = 2\sqrt{u} + C_1$

Substitute back $u = 4x - x^2$:

$I_1 = 2\sqrt{4x - x^2} + C_1$

Evaluate $I_2$:

$I_2 = \int \frac{1}{\sqrt{4x - x^2}} dx$. We complete the square for the quadratic expression in the denominator.

$4x - x^2 = -(x^2 - 4x)$

$= -((x^2 - 4x + 4) - 4)$

$= -((x - 2)^2 - 4)$

$= 4 - (x - 2)^2$

$= 2^2 - (x - 2)^2$

So, $I_2 = \int \frac{1}{\sqrt{2^2 - (x - 2)^2}} dx$.

This integral is in the form $\int \frac{1}{\sqrt{a^2 - u^2}} du$, where $u = x-2$ and $a=2$. Note that $du = dx$ for $u=x-2$.

Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \arcsin\left(\frac{x}{a}\right) + C$:

$I_2 = \arcsin\left(\frac{x - 2}{2}\right) + C_2$

Combine $I_1$ and $I_2$ to find the total integral $I = -\frac{1}{2} I_1 + 4 I_2$:

$I = -\frac{1}{2} \left(2\sqrt{4x - x^2} + C_1\right) + 4 \left(\arcsin\left(\frac{x - 2}{2}\right) + C_2\right)$

$I = -\sqrt{4x - x^2} - \frac{1}{2}C_1 + 4 \arcsin\left(\frac{x - 2}{2}\right) + 4C_2$

Let $C = -\frac{1}{2}C_1 + 4C_2$ be the constant of integration.

$I = -\sqrt{4x - x^2} + 4 \arcsin\left(\frac{x - 2}{2}\right) + C$

Thus, the integral of the given function is $-\sqrt{4x - x^2} + 4 \arcsin\left(\frac{x - 2}{2}\right) + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{x + 2}{\sqrt{x^2 + 2x + 3}}$.

Let the integral be $I$.

$I = \int \frac{x + 2}{\sqrt{x^2 + 2x + 3}} dx$

We express the numerator $x+2$ in terms of the derivative of the expression inside the square root in the denominator, $x^2 + 2x + 3$.

The derivative of $x^2 + 2x + 3$ is $2x + 2$.

We set the numerator equal to $A$ times the derivative plus a constant $B$:

$x + 2 = A(2x + 2) + B$

Expand the right side:

$x + 2 = 2Ax + 2A + B$

Equating the coefficients of $x$ on both sides:

$1 = 2A$

Solving for $A$:

$A = \frac{1}{2}$

Equating the constant terms on both sides:

$2 = 2A + B$

Substitute the value of $A = \frac{1}{2}$:

$2 = 2\left(\frac{1}{2}\right) + B$

$2 = 1 + B$

Solving for $B$:

$B = 2 - 1 = 1$

So, the numerator $x + 2$ can be written as $\frac{1}{2}(2x + 2) + 1$.

Substitute this into the integral:

$I = \int \frac{\frac{1}{2}(2x + 2) + 1}{\sqrt{x^2 + 2x + 3}} dx$

Split the integral into two separate integrals:

$I = \int \frac{\frac{1}{2}(2x + 2)}{\sqrt{x^2 + 2x + 3}} dx + \int \frac{1}{\sqrt{x^2 + 2x + 3}} dx$

$I = \frac{1}{2} \int \frac{2x + 2}{\sqrt{x^2 + 2x + 3}} dx + \int \frac{1}{\sqrt{x^2 + 2x + 3}} dx$

Let the first integral be $I_1 = \int \frac{2x + 2}{\sqrt{x^2 + 2x + 3}} dx$.

For $I_1$, let $u = x^2 + 2x + 3$. Then $du = (2x + 2) dx$.

$I_1 = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du$

Integrate $u^{-1/2}$ using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

$I_1 = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_1 = \frac{u^{1/2}}{1/2} + C_1 = 2\sqrt{u} + C_1$

Substitute back $u = x^2 + 2x + 3$:

$I_1 = 2\sqrt{x^2 + 2x + 3} + C_1$

Let the second integral be $I_2 = \int \frac{1}{\sqrt{x^2 + 2x + 3}} dx$.

To evaluate $I_2$, we complete the square for the quadratic expression in the denominator, $x^2 + 2x + 3$.

$x^2 + 2x + 3 = (x^2 + 2x + 1) - 1 + 3$

$x^2 + 2x + 3 = (x+1)^2 + 2$

$x^2 + 2x + 3 = (x+1)^2 + (\sqrt{2})^2$

Substitute the completed square form into $I_2$:

$I_2 = \int \frac{1}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} dx$

This integral is in the standard form $\int \frac{1}{\sqrt{v^2 + a^2}} dv = \ln|v + \sqrt{v^2 + a^2}| + C$.

Here, the variable is $x+1$ and $a = \sqrt{2}$.

$I_2 = \ln|(x+1) + \sqrt{(x+1)^2 + (\sqrt{2})^2}| + C_2$

Substitute back the original quadratic expression under the square root:

$I_2 = \ln|x+1 + \sqrt{x^2 + 2x + 3}| + C_2$

Now combine the results for $I_1$ and $I_2$ to get the total integral $I = \frac{1}{2} I_1 + I_2$:

$I = \frac{1}{2} (2\sqrt{x^2 + 2x + 3} + C_1) + (\ln|x+1 + \sqrt{x^2 + 2x + 3}| + C_2)$

$I = \sqrt{x^2 + 2x + 3} + \frac{1}{2}C_1 + \ln|x+1 + \sqrt{x^2 + 2x + 3}| + C_2$

Let $C = \frac{1}{2}C_1 + C_2$ be the constant of integration.

$I = \sqrt{x^2 + 2x + 3} + \ln|x+1 + \sqrt{x^2 + 2x + 3}| + C$

Thus, the integral of the given function is $\sqrt{x^2 + 2x + 3} + \ln|x+1 + \sqrt{x^2 + 2x + 3}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{x + 3}{x^2 − 2x − 5}$.

Let the integral be $I$.

$I = \int \frac{x + 3}{x^2 − 2x − 5} dx$

This is an integral of the form $\int \frac{Px+Q}{ax^2+bx+c} dx$. We express the numerator $x+3$ as $A \cdot \frac{d}{dx}(x^2-2x-5) + B$.

The derivative of the denominator $x^2 - 2x - 5$ is $2x - 2$.

We set up the equation:

$x + 3 = A(2x - 2) + B$

Expand the right side:

$x + 3 = 2Ax - 2A + B$

Equating the coefficients of $x$ on both sides:

$1 = 2A$

$A = \frac{1}{2}$

Equating the constant terms on both sides:

$3 = -2A + B$

Substitute the value of $A = \frac{1}{2}$:

$3 = -2\left(\frac{1}{2}\right) + B$

$3 = -1 + B$

$B = 3 + 1 = 4$

So, the numerator $x + 3$ can be written as $\frac{1}{2}(2x - 2) + 4$.

Substitute this expression for the numerator back into the integral:

$I = \int \frac{\frac{1}{2}(2x - 2) + 4}{x^2 - 2x - 5} dx$

Split the integral into two separate integrals:

$I = \int \frac{\frac{1}{2}(2x - 2)}{x^2 - 2x - 5} dx + \int \frac{4}{x^2 - 2x - 5} dx$

$I = \frac{1}{2} \int \frac{2x - 2}{x^2 - 2x - 5} dx + 4 \int \frac{1}{x^2 - 2x - 5} dx$

Let the first integral be $I_1 = \int \frac{2x - 2}{x^2 - 2x - 5} dx$.

For $I_1$, let $u = x^2 - 2x - 5$. Then $du = (2x - 2) dx$.

$I_1 = \int \frac{1}{u} du = \ln|u| + C_1$

Substitute back $u = x^2 - 2x - 5$:

$I_1 = \ln|x^2 - 2x - 5| + C_1$

Let the second integral be $I_2 = \int \frac{1}{x^2 - 2x - 5} dx$.

To evaluate $I_2$, we complete the square for the quadratic expression in the denominator, $x^2 - 2x - 5$.

$x^2 - 2x - 5 = (x^2 - 2x + 1) - 1 - 5$

$x^2 - 2x - 5 = (x - 1)^2 - 6$

$x^2 - 2x - 5 = (x - 1)^2 - (\sqrt{6})^2$

Substitute the completed square form into $I_2$:

$I_2 = \int \frac{1}{(x - 1)^2 - (\sqrt{6})^2} dx$

This integral is in the standard form $\int \frac{1}{v^2 - a^2} dv = \frac{1}{2a} \ln\left|\frac{v-a}{v+a}\right| + C$.

Here, the variable is $x-1$ (let $v = x-1$, so $dv=dx$) and $a = \sqrt{6}$.

$I_2 = \frac{1}{2\sqrt{6}} \ln\left|\frac{(x-1) - \sqrt{6}}{(x-1) + \sqrt{6}}\right| + C_2$

$I_2 = \frac{1}{2\sqrt{6}} \ln\left|\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}\right| + C_2$

Now combine the results for $I_1$ and $I_2$ to get the total integral $I = \frac{1}{2} I_1 + 4 I_2$:

$I = \frac{1}{2} (\ln|x^2 - 2x - 5| + C_1) + 4 \left(\frac{1}{2\sqrt{6}} \ln\left|\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}\right| + C_2\right)$

$I = \frac{1}{2} \ln|x^2 - 2x - 5| + \frac{4}{2\sqrt{6}} \ln\left|\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}\right| + \left(\frac{1}{2}C_1 + 4C_2\right)$

$I = \frac{1}{2} \ln|x^2 - 2x - 5| + \frac{2}{\sqrt{6}} \ln\left|\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}\right| + C$

Rationalize the denominator $\frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}$.

$I = \frac{1}{2} \ln|x^2 - 2x - 5| + \frac{\sqrt{6}}{3} \ln\left|\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}\right| + C$

(Here, $C$ represents the combined constant of integration).

Thus, the integral of the given function is $\frac{1}{2} \ln|x^2 - 2x - 5| + \frac{\sqrt{6}}{3} \ln\left|\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}\right| + C$, where $C$ is the constant of integration.

Solution:

We are asked to integrate the function $\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}$.

Let the integral be $I$.

$I = \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx$

This is an integral of the form $\int \frac{Px+Q}{\sqrt{ax^2+bx+c}} dx$. We express the numerator $5x+3$ in terms of the derivative of the expression inside the square root, $x^2 + 4x + 10$.

The derivative of $x^2 + 4x + 10$ is $\frac{d}{dx}(x^2 + 4x + 10) = 2x + 4$.

We write the numerator $5x + 3$ as $A \cdot (2x + 4) + B$ for some constants $A$ and $B$.

$5x + 3 = 2Ax + 4A + B$

Equating the coefficients of $x$ on both sides:

$2A = 5$

$A = \frac{5}{2}$

Equating the constant terms on both sides:

$4A + B = 3$

Substitute the value of $A = \frac{5}{2}$:

$4\left(\frac{5}{2}\right) + B = 3$

$10 + B = 3$

$B = 3 - 10 = -7$

So, the numerator $5x + 3$ can be written as $\frac{5}{2}(2x + 4) - 7$.

Substitute this expression for the numerator back into the integral:

$I = \int \frac{\frac{5}{2}(2x + 4) - 7}{\sqrt{x^2 + 4x + 10}} dx$

Split the integral into two separate integrals:

$I = \int \frac{\frac{5}{2}(2x + 4)}{\sqrt{x^2 + 4x + 10}} dx + \int \frac{-7}{\sqrt{x^2 + 4x + 10}} dx$

$I = \frac{5}{2} \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx - 7 \int \frac{1}{\sqrt{x^2 + 4x + 10}} dx$

Let the first integral be $I_1 = \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx$.

For $I_1$, let $u = x^2 + 4x + 10$. Then $du = (2x + 4) dx$.

$I_1 = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du$

Integrate $u^{-1/2}$ using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

$I_1 = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_1 = \frac{u^{1/2}}{1/2} + C_1 = 2\sqrt{u} + C_1$

Substitute back $u = x^2 + 4x + 10$:

$I_1 = 2\sqrt{x^2 + 4x + 10} + C_1$

Let the second integral be $I_2 = \int \frac{1}{\sqrt{x^2 + 4x + 10}} dx$.

To evaluate $I_2$, we complete the square for the quadratic expression in the denominator, $x^2 + 4x + 10$.

$x^2 + 4x + 10 = (x^2 + 4x + 4) - 4 + 10$

$x^2 + 4x + 10 = (x+2)^2 + 6$

$x^2 + 4x + 10 = (x+2)^2 + (\sqrt{6})^2$

Substitute the completed square form into $I_2$:

$I_2 = \int \frac{1}{\sqrt{(x+2)^2 + (\sqrt{6})^2}} dx$

This integral is in the standard form $\int \frac{1}{\sqrt{v^2 + a^2}} dv = \ln|v + \sqrt{v^2 + a^2}| + C$.

Here, the variable is $x+2$ (let $v = x+2$, so $dv=dx$) and $a = \sqrt{6}$.

$I_2 = \ln|(x+2) + \sqrt{(x+2)^2 + (\sqrt{6})^2}| + C_2$

Substitute back the original quadratic expression under the square root:

$I_2 = \ln|x+2 + \sqrt{x^2 + 4x + 10}| + C_2$

Now combine the results for $I_1$ and $I_2$ to get the total integral $I = \frac{5}{2} I_1 - 7 I_2$:

$I = \frac{5}{2} (2\sqrt{x^2 + 4x + 10} + C_1) - 7 (\ln|x+2 + \sqrt{x^2 + 4x + 10}| + C_2)$

$I = 5\sqrt{x^2 + 4x + 10} + \frac{5}{2}C_1 - 7 \ln|x+2 + \sqrt{x^2 + 4x + 10}| - 7C_2$

Let $C = \frac{5}{2}C_1 - 7C_2$ be the constant of integration.

$I = 5\sqrt{x^2 + 4x + 10} - 7 \ln|x+2 + \sqrt{x^2 + 4x + 10}| + C$

Thus, the integral of the given function is $5\sqrt{x^2 + 4x + 10} - 7 \ln|x+2 + \sqrt{x^2 + 4x + 10}| + C$, where $C$ is the constant of integration.

Solution:

We are asked to evaluate the integral $\int \frac{dx}{x^2 + 2x + 2}$.

Let the integral be $I$.

$I = \int \frac{1}{x^2 + 2x + 2} dx$

The denominator is a quadratic expression. We complete the square for $x^2 + 2x + 2$.

$x^2 + 2x + 2 = (x^2 + 2x + 1) - 1 + 2$

$x^2 + 2x + 2 = (x+1)^2 + 1$

Substitute the completed square form into the integral:

$I = \int \frac{1}{(x+1)^2 + 1^2} dx$

This integral is in the standard form $\int \frac{1}{v^2 + a^2} dv$. We can use a substitution.

Let $v = x+1$.

Differentiating both sides with respect to $x$, we get:

$\frac{dv}{dx} = \frac{d}{dx}(x+1) = 1$

So, $dv = dx$.

Now substitute $v$ and $dx$ into the integral $I$:

$I = \int \frac{1}{v^2 + 1^2} dv$

We use the standard integral formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$.

Here, the variable is $v$ and $a=1$. So, integrating with respect to $v$, we get:

$I = \frac{1}{1} \arctan\left(\frac{v}{1}\right) + C'$

$I = \arctan(v) + C'$

Substitute back $v = x+1$:

$I = \arctan(x+1) + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Comparing this result with the given options:

(A) x tan^–1 (x + 1) + C

(B) tan^–1 (x + 1) + C

(C) (x + 1) tan^–1 x + C

(D) tan^–1 x + C

The result matches option (B).

The correct answer is (B) tan^–1 (x + 1) + C.

Solution:

We are asked to evaluate the integral $I = \int \frac{dx}{\sqrt{9x − 4x^2}}$.

The expression under the square root is a quadratic in $x$: $9x - 4x^2$. We complete the square for this expression.

$9x - 4x^2 = -4x^2 + 9x$

Factor out the coefficient of $x^2$:

$-4(x^2 - \frac{9}{4}x)$

To complete the square for $x^2 - \frac{9}{4}x$, we add and subtract the square of half the coefficient of $x$. Half of $-\frac{9}{4}$ is $-\frac{9}{8}$, and its square is $\left(-\frac{9}{8}\right)^2 = \frac{81}{64}$.

$-4\left(x^2 - \frac{9}{4}x + \frac{81}{64} - \frac{81}{64}\right)$

$-4\left(\left(x - \frac{9}{8}\right)^2 - \frac{81}{64}\right)$

Distribute the $-4$:

$-4\left(x - \frac{9}{8}\right)^2 + 4 \cdot \frac{81}{64}$

$= -4\left(x - \frac{9}{8}\right)^2 + \frac{81}{16}$

Rewrite in the form $a^2 - u^2$:

$= \frac{81}{16} - 4\left(x - \frac{9}{8}\right)^2$

$= \left(\frac{9}{4}\right)^2 - 4\left(\frac{8x - 9}{8}\right)^2$

$= \left(\frac{9}{4}\right)^2 - 4\frac{(8x - 9)^2}{64}$

$= \left(\frac{9}{4}\right)^2 - \frac{(8x - 9)^2}{16}$

Substitute this back into the integral:

$I = \int \frac{dx}{\sqrt{\frac{81}{16} - \frac{(8x - 9)^2}{16}}}$

$I = \int \frac{dx}{\sqrt{\frac{1}{16} \left(81 - (8x - 9)^2\right)}}$

$I = \int \frac{dx}{\frac{1}{4}\sqrt{81 - (8x - 9)^2}}$

$I = 4 \int \frac{dx}{\sqrt{9^2 - (8x - 9)^2}}$

Let $u = 8x - 9$.

Differentiating with respect to $x$:

$\frac{du}{dx} = 8$

So, $du = 8 dx$, which means $dx = \frac{1}{8} du$.

Substitute $u$ and $dx$ into the integral:

$I = 4 \int \frac{\frac{1}{8} du}{\sqrt{9^2 - u^2}}$

$I = \frac{4}{8} \int \frac{du}{\sqrt{9^2 - u^2}}$

$I = \frac{1}{2} \int \frac{1}{\sqrt{9^2 - u^2}} du$

This is a standard integral of the form $\int \frac{1}{\sqrt{a^2 - v^2}} dv = \arcsin\left(\frac{v}{a}\right) + C$.

Here, the variable is $u$ and $a=9$.

$I = \frac{1}{2} \arcsin\left(\frac{u}{9}\right) + C'$

Substitute back $u = 8x - 9$:

$I = \frac{1}{2} \arcsin\left(\frac{8x - 9}{9}\right) + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Comparing our result with the given options:

(A) $\frac{1}{9} \sin^{-1} \left( \frac{9x − 8}{8} \right) + C$

(B) $\frac{1}{2} \sin^{-1} \left( \frac{8x − 9}{9} \right) + C$

(C) $\frac{1}{3} \sin^{-1} \left( \frac{9x − 8}{8} \right) + C$

(D) $\frac{1}{2} \sin^{-1} \left( \frac{9x − 8}{9} \right) + C$

Our result matches option (B).

The correct answer is (B) $\frac{1}{2} \sin^{-1} \left( \frac{8x − 9}{9} \right) + C$.

Solution:

We are asked to find the integral of the function $\frac{1}{(x + 1)(x + 2)}$.

Let the integral be $I$.

$I = \int \frac{1}{(x + 1)(x + 2)} dx$

The integrand is a rational function with a denominator that is a product of distinct linear factors. We can use the method of partial fraction decomposition.

Assume the integrand can be written as:

$\frac{1}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$

To find the constants $A$ and $B$, we combine the terms on the right side:

$\frac{1}{(x + 1)(x + 2)} = \frac{A(x + 2) + B(x + 1)}{(x + 1)(x + 2)}$

Equating the numerators:

$1 = A(x + 2) + B(x + 1)$

This equation must hold for all values of $x$. We can find $A$ and $B$ by substituting specific values of $x$ or by equating coefficients.

Using the substitution method:

Set $x = -1$:

$1 = A(-1 + 2) + B(-1 + 1)$

$1 = A(1) + B(0)$

$A = 1$

Set $x = -2$:

$1 = A(-2 + 2) + B(-2 + 1)$

$1 = A(0) + B(-1)$

$1 = -B$

$B = -1$

So, the partial fraction decomposition is:

$\frac{1}{(x + 1)(x + 2)} = \frac{1}{x + 1} - \frac{1}{x + 2}$

Now, we can rewrite the integral using the partial fractions:

$I = \int \left(\frac{1}{x + 1} - \frac{1}{x + 2}\right) dx$

We can split this into two separate integrals:

$I = \int \frac{1}{x + 1} dx - \int \frac{1}{x + 2} dx$

We know that $\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C$.

Applying this formula to each integral:

$\int \frac{1}{x + 1} dx = \ln|x + 1| + C_1$

$\int \frac{1}{x + 2} dx = \ln|x + 2| + C_2$

Combine the results:

$I = \ln|x + 1| - \ln|x + 2| + (C_1 - C_2)$

Let $C = C_1 - C_2$ be the constant of integration.

$I = \ln|x + 1| - \ln|x + 2| + C$

Using the property of logarithms $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$:

$I = \ln\left|\frac{x + 1}{x + 2}\right| + C$

Thus, the integral of the given function is $\ln\left|\frac{x + 1}{x + 2}\right| + C$, where $C$ is the constant of integration.

Solution:

We are asked to find the integral of the function $\frac{x^2 + 1}{x^2 − 5x + 6}$.

Let the integral be $I$.

$I = \int \frac{x^2 + 1}{x^2 − 5x + 6} dx$

The degree of the numerator ($2$) is equal to the degree of the denominator ($2$). Therefore, we first perform polynomial long division.

Divide $x^2 + 1$ by $x^2 - 5x + 6$:

$\frac{x^2 + 1}{x^2 - 5x + 6} = 1 + \frac{(x^2 + 1) - 1(x^2 - 5x + 6)}{x^2 - 5x + 6}$

$= 1 + \frac{x^2 + 1 - x^2 + 5x - 6}{x^2 - 5x + 6}$

$= 1 + \frac{5x - 5}{x^2 - 5x + 6}$

Now the integral becomes:

$I = \int \left(1 + \frac{5x - 5}{x^2 - 5x + 6}\right) dx$

$I = \int 1 \, dx + \int \frac{5x - 5}{x^2 - 5x + 6} dx$

The first part is easy to integrate: $\int 1 \, dx = x + C_1$.

For the second part, $\int \frac{5x - 5}{x^2 - 5x + 6} dx$, we use partial fraction decomposition.

First, factor the denominator: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

So, we decompose $\frac{5x - 5}{(x - 2)(x - 3)}$ into partial fractions:

$\frac{5x - 5}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$

Multiply by $(x - 2)(x - 3)$ to clear the denominators:

$5x - 5 = A(x - 3) + B(x - 2)$

To find $A$, set $x = 2$:

$5(2) - 5 = A(2 - 3) + B(2 - 2)$

$10 - 5 = A(-1) + B(0)$

$5 = -A \implies A = -5$

To find $B$, set $x = 3$:

$5(3) - 5 = A(3 - 3) + B(3 - 2)$

$15 - 5 = A(0) + B(1)$

$10 = B$

So, the partial fraction decomposition is:

$\frac{5x - 5}{x^2 - 5x + 6} = \frac{-5}{x - 2} + \frac{10}{x - 3}$

Now, integrate the second part:

$\int \frac{5x - 5}{x^2 - 5x + 6} dx = \int \left(\frac{-5}{x - 2} + \frac{10}{x - 3}\right) dx$

$= -5 \int \frac{1}{x - 2} dx + 10 \int \frac{1}{x - 3} dx$

$= -5 \ln|x - 2| + 10 \ln|x - 3| + C_2$

Combine the results of both parts of the integral:

$I = (x + C_1) + (-5 \ln|x - 2| + 10 \ln|x - 3| + C_2)$

$I = x - 5 \ln|x - 2| + 10 \ln|x - 3| + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

$I = x - 5 \ln|x - 2| + 10 \ln|x - 3| + C$

Thus, the integral is $x - 5 \ln|x - 2| + 10 \ln|x - 3| + C$, where $C$ is the constant of integration.

Solution:

We are asked to find the integral of the function $\frac{3x − 2}{(x + 1)^2 (x + 3)}$.

Let the integral be $I$.

$I = \int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx$

The integrand is a rational function with a repeated linear factor $(x+1)^2$ and a distinct linear factor $(x+3)$ in the denominator. We use the method of partial fraction decomposition.

We assume the integrand can be written in the form:

$\frac{3x − 2}{(x + 1)^2 (x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + 3}$

To find the constants $A$, $B$, and $C$, we multiply both sides by the denominator $(x + 1)^2 (x + 3)$:

$3x - 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)^2$

We can find the constants by substituting specific values of $x$ that make terms zero.

Substitute $x = -1$:

$3(-1) - 2 = A(-1 + 1)(-1 + 3) + B(-1 + 3) + C(-1 + 1)^2$

$-3 - 2 = A(0)(2) + B(2) + C(0)^2$

$-5 = 2B$

$B = -\frac{5}{2}$

Substitute $x = -3$:

$3(-3) - 2 = A(-3 + 1)(-3 + 3) + B(-3 + 3) + C(-3 + 1)^2$

$-9 - 2 = A(-2)(0) + B(0) + C(-2)^2$

$-11 = 4C$

$C = -\frac{11}{4}$

To find $A$, we can substitute another value for $x$, for example $x = 0$, and use the values of $B$ and $C$ we just found.

Substitute $x = 0$:

$3(0) - 2 = A(0 + 1)(0 + 3) + B(0 + 3) + C(0 + 1)^2$

$-2 = A(1)(3) + B(3) + C(1)^2$

$-2 = 3A + 3B + C$

Substitute $B = -\frac{5}{2}$ and $C = -\frac{11}{4}$ into this equation:

$-2 = 3A + 3\left(-\frac{5}{2}\right) + \left(-\frac{11}{4}\right)$

$-2 = 3A - \frac{15}{2} - \frac{11}{4}$

To solve for $3A$, move the other terms to the left side:

$3A = -2 + \frac{15}{2} + \frac{11}{4}$

Find a common denominator, which is 4:

$3A = -\frac{8}{4} + \frac{30}{4} + \frac{11}{4}$

$3A = \frac{-8 + 30 + 11}{4}$

$3A = \frac{22 + 11}{4}$

$3A = \frac{33}{4}$

$A = \frac{33}{4} \cdot \frac{1}{3}$

$A = \frac{11}{4}$

So, the partial fraction decomposition is:

$\frac{3x - 2}{(x + 1)^2 (x + 3)} = \frac{11/4}{x + 1} + \frac{-5/2}{(x + 1)^2} + \frac{-11/4}{x + 3}$

$\frac{3x - 2}{(x + 1)^2 (x + 3)} = \frac{11}{4(x + 1)} - \frac{5}{2(x + 1)^2} - \frac{11}{4(x + 3)}$

Now we can integrate the expression term by term:

$I = \int \left(\frac{11}{4(x + 1)} - \frac{5}{2(x + 1)^2} - \frac{11}{4(x + 3)}\right) dx$

$I = \frac{11}{4} \int \frac{1}{x + 1} dx - \frac{5}{2} \int \frac{1}{(x + 1)^2} dx - \frac{11}{4} \int \frac{1}{x + 3} dx$

Evaluate each integral:

$\int \frac{1}{x + 1} dx = \ln|x + 1|$

$\int \frac{1}{(x + 1)^2} dx = \int (x + 1)^{-2} dx$. Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$ with $u = x+1$ and $n = -2$:

$\int (x + 1)^{-2} dx = \frac{(x + 1)^{-2 + 1}}{-2 + 1} = \frac{(x + 1)^{-1}}{-1} = -\frac{1}{x + 1}$

$\int \frac{1}{x + 3} dx = \ln|x + 3|$

Substitute these results back into the integral expression for $I$:

$I = \frac{11}{4} (\ln|x + 1|) - \frac{5}{2} \left(-\frac{1}{x + 1}\right) - \frac{11}{4} (\ln|x + 3|) + C$

where $C$ is the constant of integration.

$I = \frac{11}{4} \ln|x + 1| + \frac{5}{2(x + 1)} - \frac{11}{4} \ln|x + 3| + C$

We can group the logarithmic terms using the logarithm property $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$:

$I = \frac{11}{4} (\ln|x + 1| - \ln|x + 3|) + \frac{5}{2(x + 1)} + C$

$I = \frac{11}{4} \ln\left|\frac{x + 1}{x + 3}\right| + \frac{5}{2(x + 1)} + C$

Thus, the integral of the given function is $\frac{11}{4} \ln\left|\frac{x + 1}{x + 3}\right| + \frac{5}{2(x + 1)} + C$, where $C$ is the constant of integration.

Solution:

We are asked to find the integral of the function $\frac{x^2}{(x^2 + 1) (x^2 + 4)}$.

Let the integral be $I$.

$I = \int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx$

The integrand is a proper rational function, and the denominator is a product of irreducible quadratic factors. We can use the method of partial fraction decomposition.

Assume the integrand can be written in the form:

$\frac{x^2}{(x^2 + 1) (x^2 + 4)} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + 4}$

To find the constants $A, B, C, D$, we multiply both sides by the denominator $(x^2 + 1)(x^2 + 4)$:

$x^2 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 + 1)$

Expand the right side and group terms by powers of $x$:

$x^2 = (Ax^3 + 4Ax + Bx^2 + 4B) + (Cx^3 + Cx + Dx^2 + D)$

$x^2 = (A + C)x^3 + (B + D)x^2 + (4A + C)x + (4B + D)$

Equating the coefficients of corresponding powers of $x$ on both sides:

Coefficient of $x^3$: $0 = A + C$

Coefficient of $x^2$: $1 = B + D$

Coefficient of $x^1$: $0 = 4A + C$

Coefficient of $x^0$: $0 = 4B + D$

From the first equation, $C = -A$. Substitute this into the third equation:

$0 = 4A + (-A)$

$0 = 3A \implies A = 0$

Since $C = -A$, we have $C = 0$.

From the second equation, $D = 1 - B$. Substitute this into the fourth equation:

$0 = 4B + (1 - B)$

$0 = 3B + 1$

$3B = -1 \implies B = -\frac{1}{3}$

Since $D = 1 - B$, we have $D = 1 - \left(-\frac{1}{3}\right) = 1 + \frac{1}{3} = \frac{4}{3}$.

So, the partial fraction decomposition is:

$\frac{x^2}{(x^2 + 1) (x^2 + 4)} = \frac{0 \cdot x + (-\frac{1}{3})}{x^2 + 1} + \frac{0 \cdot x + \frac{4}{3}}{x^2 + 4}$

$\frac{x^2}{(x^2 + 1) (x^2 + 4)} = -\frac{1}{3(x^2 + 1)} + \frac{4}{3(x^2 + 4)}$

Now we integrate the decomposed form:

$I = \int \left(-\frac{1}{3(x^2 + 1)} + \frac{4}{3(x^2 + 4)}\right) dx$

$I = -\frac{1}{3} \int \frac{1}{x^2 + 1^2} dx + \frac{4}{3} \int \frac{1}{x^2 + 2^2} dx$

Using the standard integral formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$:

The first integral is $-\frac{1}{3} \cdot \frac{1}{1} \arctan\left(\frac{x}{1}\right) + C_1 = -\frac{1}{3} \arctan(x) + C_1$.

The second integral is $\frac{4}{3} \cdot \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C_2 = \frac{2}{3} \arctan\left(\frac{x}{2}\right) + C_2$.

Combining the results:

$I = -\frac{1}{3} \arctan(x) + \frac{2}{3} \arctan\left(\frac{x}{2}\right) + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

$I = \frac{2}{3} \arctan\left(\frac{x}{2}\right) - \frac{1}{3} \arctan(x) + C$

Thus, the integral of the given function is $\frac{2}{3} \arctan\left(\frac{x}{2}\right) - \frac{1}{3} \arctan(x) + C$, where $C$ is the constant of integration.

Solution:

We are asked to find the integral $I = \int \frac{(3\sin φ − 2) \cos φ}{5 − \cos^2 φ − 4 \sin φ} \;dφ$.

Let's simplify the denominator using the identity $\cos^2 φ = 1 - \sin^2 φ$.

Denominator $= 5 - (1 - \sin^2 φ) - 4 \sin φ$

$= 5 - 1 + \sin^2 φ - 4 \sin φ$

$= \sin^2 φ - 4 \sin φ + 4$

$= (\sin φ - 2)^2$

So the integral becomes $I = \int \frac{(3\sin φ − 2) \cos φ}{(\sin φ - 2)^2} \;dφ$.

Let's use the substitution method. Let $u = \sin φ$.

Differentiating with respect to $φ$, we get $\frac{du}{dφ} = \cos φ$, so $du = \cos φ \;dφ$.

Substitute $u = \sin φ$ and $du = \cos φ \;dφ$ into the integral:

$I = \int \frac{(3u - 2)}{(u - 2)^2} \;du$

The integrand is a rational function. We use the method of partial fraction decomposition.

Assume $\frac{3u - 2}{(u - 2)^2} = \frac{A}{u - 2} + \frac{B}{(u - 2)^2}$.

Multiply both sides by $(u - 2)^2$:

$3u - 2 = A(u - 2) + B$

$3u - 2 = Au - 2A + B$

Equating coefficients of $u$:

$3 = A$

Equating constant terms:

$-2 = -2A + B$

Substitute $A=3$:

$-2 = -2(3) + B$

$-2 = -6 + B$

$B = -2 + 6 = 4$

So, the partial fraction decomposition is:

$\frac{3u - 2}{(u - 2)^2} = \frac{3}{u - 2} + \frac{4}{(u - 2)^2}$

Now we integrate with respect to $u$:

$I = \int \left(\frac{3}{u - 2} + \frac{4}{(u - 2)^2}\right) du$

$I = 3 \int \frac{1}{u - 2} du + 4 \int (u - 2)^{-2} du$

Using the standard integral formulas $\int \frac{1}{x} dx = \ln|x|$ and $\int x^n dx = \frac{x^{n+1}}{n+1}$:

$\int \frac{1}{u - 2} du = \ln|u - 2|$

$\int (u - 2)^{-2} du = \frac{(u - 2)^{-1}}{-1} = -\frac{1}{u - 2}$

So, the integral in terms of $u$ is:

$I = 3 \ln|u - 2| + 4 \left(-\frac{1}{u - 2}\right) + C'$

$I = 3 \ln|u - 2| - \frac{4}{u - 2} + C'$

Substitute back $u = \sin φ$:

$I = 3 \ln|\sin φ - 2| - \frac{4}{\sin φ - 2} + C$

(Here, $C$ replaces $C'$ as the constant of integration).

Since $-1 \leq \sin φ \leq 1$, $\sin φ - 2$ is always negative ($ -3 \leq \sin φ - 2 \leq -1 $). Therefore, $|\sin φ - 2| = -(\sin φ - 2) = 2 - \sin φ$. Also, $\frac{1}{\sin φ - 2} = -\frac{1}{2 - \sin φ}$.

$I = 3 \ln(2 - \sin φ) - 4 \left(-\frac{1}{2 - \sin φ}\right) + C$

$I = 3 \ln(2 - \sin φ) + \frac{4}{2 - \sin φ} + C$

Thus, the integral of the given function is $3 \ln(2 - \sin φ) + \frac{4}{2 - \sin φ} + C$, where $C$ is the constant of integration.

Solution:

We are asked to find the integral $I = \int \frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} \;dx$.

The integrand is a proper rational function with a denominator containing a linear factor $(x+2)$ and an irreducible quadratic factor $(x^2+1)$. We use partial fraction decomposition.

We assume the integrand can be written in the form:

$\frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1}$

To find the constants $A$, $B$, and $C$, we multiply both sides by the denominator $(x + 2)(x^2 + 1)$:

$x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 2)$

Expand the right side:

$x^2 + x + 1 = Ax^2 + A + Bx^2 + 2Bx + Cx + 2C$

Group terms by powers of $x$:

$x^2 + x + 1 = (A + B)x^2 + (2B + C)x + (A + 2C)$

Equating the coefficients of corresponding powers of $x$ on both sides:

Coefficient of $x^2$: $1 = A + B$

Coefficient of $x$: $1 = 2B + C$

Constant term: $1 = A + 2C$

Alternatively, we can use a combination of substitution and equating coefficients.

Substitute $x = -2$ into the equation $x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 2)$:

$(-2)^2 + (-2) + 1 = A((-2)^2 + 1) + (B(-2) + C)(-2 + 2)$

$4 - 2 + 1 = A(4 + 1) + (-2B + C)(0)$

$3 = 5A$

$A = \frac{3}{5}$

Now use the equations from equating coefficients. From $1 = A + B$, substitute $A = \frac{3}{5}$:

$1 = \frac{3}{5} + B$

$B = 1 - \frac{3}{5} = \frac{5}{5} - \frac{3}{5} = \frac{2}{5}$

From $1 = A + 2C$, substitute $A = \frac{3}{5}$:

$1 = \frac{3}{5} + 2C$

$2C = 1 - \frac{3}{5} = \frac{2}{5}$

$C = \frac{1}{5}$

So, the partial fraction decomposition is:

$\frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} = \frac{\frac{3}{5}}{x + 2} + \frac{\frac{2}{5}x + \frac{1}{5}}{x^2 + 1}$

$\frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} = \frac{3}{5(x + 2)} + \frac{2x + 1}{5(x^2 + 1)}$

Now, we can integrate the decomposed form:

$I = \int \left(\frac{3}{5(x + 2)} + \frac{2x + 1}{5(x^2 + 1)}\right) dx$

$I = \frac{3}{5} \int \frac{1}{x + 2} dx + \frac{1}{5} \int \frac{2x + 1}{x^2 + 1} dx$

Split the second integral in the sum:

$I = \frac{3}{5} \int \frac{1}{x + 2} dx + \frac{1}{5} \int \frac{2x}{x^2 + 1} dx + \frac{1}{5} \int \frac{1}{x^2 + 1} dx$

Evaluate each integral:

$\int \frac{1}{x + 2} dx = \ln|x + 2|$

For $\int \frac{2x}{x^2 + 1} dx$, let $u = x^2 + 1$, then $du = 2x dx$.

$\int \frac{2x}{x^2 + 1} dx = \int \frac{1}{u} du = \ln|u| + C_1 = \ln|x^2 + 1| + C_1$

Since $x^2 + 1$ is always positive, $|x^2 + 1| = x^2 + 1$.

$\int \frac{2x}{x^2 + 1} dx = \ln(x^2 + 1) + C_1$

$\int \frac{1}{x^2 + 1} dx$. This is a standard integral of the form $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right)$. Here $a=1$.

$\int \frac{1}{x^2 + 1} dx = \arctan(x) + C_2$

Combine the results:

$I = \frac{3}{5} \ln|x + 2| + \frac{1}{5} (\ln(x^2 + 1) + C_1) + \frac{1}{5} (\arctan(x) + C_2) + C_3$ (from the first integral)

$I = \frac{3}{5} \ln|x + 2| + \frac{1}{5} \ln(x^2 + 1) + \frac{1}{5} \arctan(x) + (\frac{1}{5}C_1 + \frac{1}{5}C_2 + C_3)$

Let $C = \frac{1}{5}C_1 + \frac{1}{5}C_2 + C_3$ be the constant of integration.

$I = \frac{3}{5} \ln|x + 2| + \frac{1}{5} \ln(x^2 + 1) + \frac{1}{5} \arctan(x) + C$

Thus, the integral of the given function is $\frac{3}{5} \ln|x + 2| + \frac{1}{5} \ln(x^2 + 1) + \frac{1}{5} \arctan(x) + C$, where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{x}{(x + 1) (x + 2)} dx$

We will use the method of partial fraction decomposition since the denominator is a product of distinct linear factors.

Let $\frac{x}{(x + 1) (x + 2)} = \frac{A}{x+1} + \frac{B}{x+2}$.

Multiplying both sides by $(x+1)(x+2)$, we get:

$x = A(x+2) + B(x+1)$

This equation must hold for all values of $x$. We can find the values of $A$ and $B$ by substituting convenient values for $x$.

Setting $x = -1$, we get:

$-1 = A(-1+2) + B(-1+1)$

$-1 = A(1) + B(0)$

$-1 = A$

Setting $x = -2$, we get:

$-2 = A(-2+2) + B(-2+1)$

$-2 = A(0) + B(-1)$

$-2 = -B$

$B = 2$

So, the partial fraction decomposition is:

$\frac{x}{(x + 1) (x + 2)} = \frac{-1}{x+1} + \frac{2}{x+2}$

Now, we can integrate the decomposed expression:

$\int \frac{x}{(x + 1) (x + 2)} dx = \int \left(\frac{-1}{x+1} + \frac{2}{x+2}\right) dx$

This integral can be split into two separate integrals:

$\int \frac{-1}{x+1} dx + \int \frac{2}{x+2} dx$

Integrating each term:

$\int \frac{-1}{x+1} dx = - \int \frac{1}{x+1} dx = -\ln|x+1| + C_1$

$\int \frac{2}{x+2} dx = 2 \int \frac{1}{x+2} dx = 2\ln|x+2| + C_2$

Combining the results and the constants of integration:

$-\ln|x+1| + 2\ln|x+2| + C$, where $C = C_1 + C_2$.

Using logarithm properties ($m\ln a = \ln a^m$ and $\ln a - \ln b = \ln \frac{a}{b}$), the result can be written as:

$2\ln|x+2| - \ln|x+1| + C = \ln|(x+2)^2| - \ln|x+1| + C = \ln\left|\frac{(x+2)^2}{x+1}\right| + C$

Thus, the integral is:

$\int \frac{x}{(x + 1) (x + 2)} dx = \textbf{$-\ln|x+1| + 2\ln|x+2| + C$}$

or

$\int \frac{x}{(x + 1) (x + 2)} dx = \textbf{$\ln\left|\frac{(x+2)^2}{x+1}\right| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{1}{x^2 - 9} dx$

We can factor the denominator as a difference of squares:

$x^2 - 9 = (x-3)(x+3)$

So the integral is:

$\int \frac{1}{(x-3)(x+3)} dx$

We will use the method of partial fraction decomposition.

Let $\frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3}$.

Multiplying both sides by $(x-3)(x+3)$, we get:

$1 = A(x+3) + B(x-3)$

To find $A$, set $x = 3$:

$1 = A(3+3) + B(3-3)$

$1 = 6A + 0$

$A = \frac{1}{6}$

To find $B$, set $x = -3$:

$1 = A(-3+3) + B(-3-3)$

$1 = 0 + B(-6)$

$1 = -6B$

$B = -\frac{1}{6}$

So, the partial fraction decomposition is:

$\frac{1}{(x-3)(x+3)} = \frac{1/6}{x-3} + \frac{-1/6}{x+3} = \frac{1}{6(x-3)} - \frac{1}{6(x+3)}$

Now, we can integrate the decomposed expression:

$\int \frac{1}{x^2 - 9} dx = \int \left(\frac{1}{6(x-3)} - \frac{1}{6(x+3)}\right) dx$

This can be written as:

$\frac{1}{6} \int \frac{1}{x-3} dx - \frac{1}{6} \int \frac{1}{x+3} dx$

Integrating each term:

$\int \frac{1}{x-3} dx = \ln|x-3| + C_1$

$\int \frac{1}{x+3} dx = \ln|x+3| + C_2$

Combining the results:

$\frac{1}{6}\ln|x-3| - \frac{1}{6}\ln|x+3| + C$, where $C = \frac{1}{6}(C_1 - C_2)$.

Using logarithm properties ($\ln a - \ln b = \ln \frac{a}{b}$), the result can be written as:

$\frac{1}{6}(\ln|x-3| - \ln|x+3|) + C = \frac{1}{6}\ln\left|\frac{x-3}{x+3}\right| + C$

Thus, the integral is:

$\int \frac{1}{x^2 - 9} dx = \textbf{$\frac{1}{6}\ln\left|\frac{x-3}{x+3}\right| + C$}$

Alternatively, we can use the standard integration formula $\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$.

In this case, $a^2 = 9$, so $a = 3$.

Using the formula directly:

$\int \frac{1}{x^2 - 9} dx = \int \frac{1}{x^2 - 3^2} dx = \frac{1}{2(3)}\ln\left|\frac{x-3}{x+3}\right| + C$

$ = \frac{1}{6}\ln\left|\frac{x-3}{x+3}\right| + C$

This matches the result obtained by partial fraction decomposition.

The final answer is $\textbf{$\frac{1}{6}\ln\left|\frac{x-3}{x+3}\right| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{3x - 1}{(x - 1) (x - 2) (x - 3)} dx$

The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

Let $\frac{3x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.

Multiplying both sides by $(x-1)(x-2)(x-3)$, we get:

$3x - 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$

We find the values of A, B, and C by substituting the roots of the denominator.

Setting $x = 1$:

$3(1) - 1 = A(1-2)(1-3) + B(1-1)(1-3) + C(1-1)(1-2)$

$2 = A(-1)(-2) + B(0)(-2) + C(0)(-1)$

$2 = 2A + 0 + 0$

$2A = 2 \implies A = 1$

Setting $x = 2$:

$3(2) - 1 = A(2-2)(2-3) + B(2-1)(2-3) + C(2-1)(2-2)$

$5 = A(0)(-1) + B(1)(-1) + C(1)(0)$

$5 = 0 - B + 0$

$5 = -B \implies B = -5$

Setting $x = 3$:

$3(3) - 1 = A(3-2)(3-3) + B(3-1)(3-3) + C(3-1)(3-2)$

$8 = A(1)(0) + B(2)(0) + C(2)(1)$

$8 = 0 + 0 + 2C$

$8 = 2C \implies C = 4$

So, the partial fraction decomposition is:

$\frac{3x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{1}{x-1} + \frac{-5}{x-2} + \frac{4}{x-3}$

$\frac{3x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3}$

Now, we integrate the decomposed expression:

$\int \frac{3x - 1}{(x - 1) (x - 2) (x - 3)} dx = \int \left(\frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3}\right) dx$

$= \int \frac{1}{x-1} dx - 5\int \frac{1}{x-2} dx + 4\int \frac{1}{x-3} dx$

Integrating each term, we get:

$\int \frac{1}{x-1} dx = \ln|x-1| + C_1$

$\int \frac{1}{x-2} dx = \ln|x-2| + C_2$

$\int \frac{1}{x-3} dx = \ln|x-3| + C_3$

Combining the results:

$\int \frac{3x - 1}{(x - 1) (x - 2) (x - 3)} dx = \ln|x-1| - 5\ln|x-2| + 4\ln|x-3| + C$

where $C = C_1 - 5C_2 + 4C_3$ is the constant of integration.

The final answer is $\textbf{$\ln|x-1| - 5\ln|x-2| + 4\ln|x-3| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} dx$

The denominator consists of distinct linear factors. We use the method of partial fraction decomposition.

Let $\frac{x}{(x - 1) (x - 2) (x - 3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.

Multiplying both sides by $(x-1)(x-2)(x-3)$, we get:

$x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$

We find the values of A, B, and C by substituting the roots of the denominator.

Setting $x = 1$:

$1 = A(1-2)(1-3) + B(1-1)(1-3) + C(1-1)(1-2)$

$1 = A(-1)(-2) + B(0)(-2) + C(0)(-1)$

$1 = 2A + 0 + 0$

$2A = 1 \implies A = \frac{1}{2}$

Setting $x = 2$:

$2 = A(2-2)(2-3) + B(2-1)(2-3) + C(2-1)(2-2)$

$2 = A(0)(-1) + B(1)(-1) + C(1)(0)$

$2 = 0 - B + 0$

$2 = -B \implies B = -2$

Setting $x = 3$:

$3 = A(3-2)(3-3) + B(3-1)(3-3) + C(3-1)(3-2)$

$3 = A(1)(0) + B(2)(0) + C(2)(1)$

$3 = 0 + 0 + 2C$

$3 = 2C \implies C = \frac{3}{2}$

So, the partial fraction decomposition is:

$\frac{x}{(x - 1) (x - 2) (x - 3)} = \frac{1/2}{x-1} + \frac{-2}{x-2} + \frac{3/2}{x-3}$

$\frac{x}{(x - 1) (x - 2) (x - 3)} = \frac{1}{2(x-1)} - \frac{2}{x-2} + \frac{3}{2(x-3)}$

Now, we integrate the decomposed expression:

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} dx = \int \left(\frac{1}{2(x-1)} - \frac{2}{x-2} + \frac{3}{2(x-3)}\right) dx$

$= \frac{1}{2}\int \frac{1}{x-1} dx - 2\int \frac{1}{x-2} dx + \frac{3}{2}\int \frac{1}{x-3} dx$

Integrating each term, we get:

$\int \frac{1}{x-1} dx = \ln|x-1| + C_1$

$\int \frac{1}{x-2} dx = \ln|x-2| + C_2$

$\int \frac{1}{x-3} dx = \ln|x-3| + C_3$

Combining the results:

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} dx = \frac{1}{2}\ln|x-1| - 2\ln|x-2| + \frac{3}{2}\ln|x-3| + C$

where $C$ is the constant of integration.

The final answer is $\textbf{$\frac{1}{2}\ln|x-1| - 2\ln|x-2| + \frac{3}{2}\ln|x-3| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{2x}{x^2 + 3x + 2} dx$

First, we factor the denominator:

$x^2 + 3x + 2 = (x+1)(x+2)$

So, the integral becomes:

$\int \frac{2x}{(x+1)(x+2)} dx$

We use the method of partial fraction decomposition.

Let $\frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.

Multiplying both sides by $(x+1)(x+2)$, we get:

$2x = A(x+2) + B(x+1)$

To find the coefficients $A$ and $B$, we substitute the roots of the denominator.

Setting $x = -1$:

$2(-1) = A(-1+2) + B(-1+1)$

$-2 = A(1) + B(0)$

$-2 = A$

Setting $x = -2$:

$2(-2) = A(-2+2) + B(-2+1)$

$-4 = A(0) + B(-1)$

$-4 = -B$

$B = 4$

So, the partial fraction decomposition is:

$\frac{2x}{(x+1)(x+2)} = \frac{-2}{x+1} + \frac{4}{x+2}$

Now, we integrate the decomposed expression:

$\int \frac{2x}{(x+1)(x+2)} dx = \int \left(\frac{-2}{x+1} + \frac{4}{x+2}\right) dx$

$= -2\int \frac{1}{x+1} dx + 4\int \frac{1}{x+2} dx$

Integrating each term, we get:

$\int \frac{1}{x+1} dx = \ln|x+1| + C_1$

$\int \frac{1}{x+2} dx = \ln|x+2| + C_2$

Combining the results:

$\int \frac{2x}{x^2 + 3x + 2} dx = -2\ln|x+1| + 4\ln|x+2| + C$

where $C = -2C_1 + 4C_2$ is the constant of integration.

Using logarithm properties, the answer can also be written as:

$4\ln|x+2| - 2\ln|x+1| + C = \ln|(x+2)^4| - \ln|(x+1)^2| + C = \ln\left|\frac{(x+2)^4}{(x+1)^2}\right| + C$

The final answer is $\textbf{$-2\ln|x+1| + 4\ln|x+2| + C$}$

or

$\textbf{$\ln\left|\frac{(x+2)^4}{(x+1)^2}\right| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{1 - x^2}{x (1 - 2x)} dx$

The rational function is $\frac{1 - x^2}{x - 2x^2}$. The degree of the numerator (2) is equal to the degree of the denominator (2). Therefore, it is an improper rational function, and we must perform polynomial long division first.

We divide $-x^2 + 1$ by $-2x^2 + x$.

$\begin{array}{r} \frac{1}{2}\phantom{+0} \\ -2x^2+x{\overline{\smash{\big)}\,-x^2+0x+1\phantom{)}}}} \\ \underline{-~\phantom{()}(-x^2+\frac{1}{2}x)} \\ -\frac{1}{2}x+1\phantom{)} \end{array}$

So, $\frac{1 - x^2}{x (1 - 2x)} = \frac{1}{2} + \frac{-\frac{1}{2}x + 1}{-2x^2 + x} = \frac{1}{2} + \frac{-\frac{1}{2}x + 1}{-x(2x - 1)} = \frac{1}{2} + \frac{\frac{1}{2}x - 1}{x(2x - 1)}$.

The integral becomes:

$\int \left(\frac{1}{2} + \frac{\frac{1}{2}x - 1}{x(2x - 1)}\right) dx = \int \frac{1}{2} dx + \int \frac{\frac{1}{2}x - 1}{x(2x - 1)} dx$

Now, we use partial fraction decomposition for the remaining fraction $\frac{\frac{1}{2}x - 1}{x(2x - 1)}$.

Let $\frac{\frac{1}{2}x - 1}{x(2x - 1)} = \frac{A}{x} + \frac{B}{2x - 1}$.

Multiplying by $x(2x - 1)$, we get:

$\frac{1}{2}x - 1 = A(2x - 1) + Bx$

Setting $x = 0$:

$\frac{1}{2}(0) - 1 = A(2(0) - 1) + B(0)$

$-1 = A(-1)$

$A = 1$

Setting $x = \frac{1}{2}$ (from $2x - 1 = 0$):

$\frac{1}{2}\left(\frac{1}{2}\right) - 1 = A\left(2\left(\frac{1}{2}\right) - 1\right) + B\left(\frac{1}{2}\right)$

$\frac{1}{4} - 1 = A(1 - 1) + \frac{1}{2}B$

$-\frac{3}{4} = 0 + \frac{1}{2}B$

$B = -\frac{3}{4} \times 2 = -\frac{3}{2}$

So, $\frac{\frac{1}{2}x - 1}{x(2x - 1)} = \frac{1}{x} + \frac{-3/2}{2x - 1} = \frac{1}{x} - \frac{3}{2(2x - 1)}$.

Now we integrate the decomposed parts:

$\int \frac{1 - x^2}{x (1 - 2x)} dx = \int \left(\frac{1}{2} + \frac{1}{x} - \frac{3}{2(2x - 1)}\right) dx$

$= \int \frac{1}{2} dx + \int \frac{1}{x} dx - \frac{3}{2}\int \frac{1}{2x - 1} dx$

Integrating each term:

$\int \frac{1}{2} dx = \frac{1}{2}x + C_1$

$\int \frac{1}{x} dx = \ln|x| + C_2$

For the third integral, let $u = 2x - 1$, so $du = 2 dx$, which means $dx = \frac{1}{2} du$.

$-\frac{3}{2}\int \frac{1}{2x - 1} dx = -\frac{3}{2}\int \frac{1}{u} \left(\frac{1}{2} du\right) = -\frac{3}{4}\int \frac{1}{u} du = -\frac{3}{4}\ln|u| + C_3 = -\frac{3}{4}\ln|2x - 1| + C_3$

Combining the results:

$\int \frac{1 - x^2}{x (1 - 2x)} dx = \frac{1}{2}x + \ln|x| - \frac{3}{4}\ln|2x - 1| + C$

where $C = C_1 + C_2 + C_3$ is the constant of integration.

The final answer is $\textbf{$\frac{1}{2}x + \ln|x| - \frac{3}{4}\ln|2x - 1| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{x}{(x^2 + 1) (x − 1)} dx$

The denominator has a linear factor $(x-1)$ and an irreducible quadratic factor $(x^2+1)$. We use the method of partial fraction decomposition.

Let $\frac{x}{(x^2 + 1) (x − 1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$.

Multiplying both sides by $(x-1)(x^2+1)$, we get:

$x = A(x^2+1) + (Bx+C)(x-1)$

To find the coefficients $A$, $B$, and $C$, we can substitute values for $x$ or compare coefficients.

Setting $x = 1$:

$1 = A(1^2+1) + (B(1)+C)(1-1)$

$1 = A(2) + (B+C)(0)$

$1 = 2A$

$A = \frac{1}{2}$

Setting $x = 0$:

$0 = A(0^2+1) + (B(0)+C)(0-1)$

$0 = A(1) + (C)(-1)$

$0 = A - C$

Since $A = \frac{1}{2}$, we have $C = \frac{1}{2}$.

Setting $x = -1$:

$-1 = A((-1)^2+1) + (B(-1)+C)(-1-1)$

$-1 = A(1+1) + (-B+C)(-2)$

$-1 = 2A + 2B - 2C$

Substitute $A = \frac{1}{2}$ and $C = \frac{1}{2}$:

$-1 = 2\left(\frac{1}{2}\right) + 2B - 2\left(\frac{1}{2}\right)$

$-1 = 1 + 2B - 1$

$-1 = 2B$

$B = -\frac{1}{2}$

So, the partial fraction decomposition is:

$\frac{x}{(x^2 + 1) (x − 1)} = \frac{1/2}{x-1} + \frac{(-1/2)x + 1/2}{x^2+1} = \frac{1}{2(x-1)} + \frac{1-x}{2(x^2+1)}$

Now, we integrate the decomposed expression:

$\int \frac{x}{(x^2 + 1) (x − 1)} dx = \int \left(\frac{1}{2(x-1)} + \frac{1-x}{2(x^2+1)}\right) dx$

$= \frac{1}{2}\int \frac{1}{x-1} dx + \frac{1}{2}\int \frac{1-x}{x^2+1} dx$

We can split the second integral:

$= \frac{1}{2}\int \frac{1}{x-1} dx + \frac{1}{2}\int \frac{1}{x^2+1} dx - \frac{1}{2}\int \frac{x}{x^2+1} dx$

Integrate each term:

$\int \frac{1}{x-1} dx = \ln|x-1| + C_1$

$\int \frac{1}{x^2+1} dx = \arctan(x) + C_2$

For $\int \frac{x}{x^2+1} dx$, let $u = x^2+1$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$.

$\int \frac{x}{x^2+1} dx = \int \frac{1}{u} \frac{1}{2} du = \frac{1}{2}\int \frac{1}{u} du = \frac{1}{2}\ln|u| + C_3 = \frac{1}{2}\ln(x^2+1) + C_3$ (since $x^2+1 > 0$).

Combining the results:

$\int \frac{x}{(x^2 + 1) (x − 1)} dx = \frac{1}{2}(\ln|x-1|) + \frac{1}{2}(\arctan(x)) - \frac{1}{2}\left(\frac{1}{2}\ln(x^2+1)\right) + C$

$= \frac{1}{2}\ln|x-1| + \frac{1}{2}\arctan(x) - \frac{1}{4}\ln(x^2+1) + C$

where $C$ is the constant of integration.

The final answer is $\textbf{$\frac{1}{2}\ln|x-1| + \frac{1}{2}\arctan(x) - \frac{1}{4}\ln(x^2+1) + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{x}{(x - 1)^2 (x + 2)} dx$

The denominator has a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x+2)$. We use the method of partial fraction decomposition.

Let $\frac{x}{(x - 1)^2 (x + 2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$.

Multiplying both sides by $(x-1)^2(x+2)$, we get:

$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$

We find the values of A, B, and C by substituting the roots of the denominator and a convenient value for $x$.

Setting $x = 1$:

$1 = A(1-1)(1+2) + B(1+2) + C(1-1)^2$

$1 = A(0)(3) + B(3) + C(0)^2$

$1 = 0 + 3B + 0$

$3B = 1 \implies B = \frac{1}{3}$

Setting $x = -2$:

$-2 = A(-2-1)(-2+2) + B(-2+2) + C(-2-1)^2$

$-2 = A(-3)(0) + B(0) + C(-3)^2$

$-2 = 0 + 0 + 9C$

$9C = -2 \implies C = -\frac{2}{9}$

Setting $x = 0$:

$0 = A(0-1)(0+2) + B(0+2) + C(0-1)^2$

$0 = A(-1)(2) + B(2) + C(-1)^2$

$0 = -2A + 2B + C$

Substitute the values of B and C:

$0 = -2A + 2\left(\frac{1}{3}\right) + \left(-\frac{2}{9}\right)$

$0 = -2A + \frac{2}{3} - \frac{2}{9}$

$0 = -2A + \frac{6}{9} - \frac{2}{9}$

$0 = -2A + \frac{4}{9}$

$2A = \frac{4}{9}$

$A = \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$

So, the partial fraction decomposition is:

$\frac{x}{(x - 1)^2 (x + 2)} = \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} + \frac{-2/9}{x+2}$

$\frac{x}{(x - 1)^2 (x + 2)} = \frac{2}{9(x-1)} + \frac{1}{3(x-1)^2} - \frac{2}{9(x+2)}$

Now, we integrate the decomposed expression:

$\int \frac{x}{(x - 1)^2 (x + 2)} dx = \int \left(\frac{2}{9(x-1)} + \frac{1}{3(x-1)^2} - \frac{2}{9(x+2)}\right) dx$

$= \frac{2}{9}\int \frac{1}{x-1} dx + \frac{1}{3}\int (x-1)^{-2} dx - \frac{2}{9}\int \frac{1}{x+2} dx$

Integrating each term:

$\int \frac{1}{x-1} dx = \ln|x-1| + C_1$

$\int (x-1)^{-2} dx = \frac{(x-1)^{-2+1}}{-2+1} + C_2 = \frac{(x-1)^{-1}}{-1} + C_2 = -\frac{1}{x-1} + C_2$

$\int \frac{1}{x+2} dx = \ln|x+2| + C_3$

Combining the results:

$\int \frac{x}{(x - 1)^2 (x + 2)} dx = \frac{2}{9}\ln|x-1| + \frac{1}{3}\left(-\frac{1}{x-1}\right) - \frac{2}{9}\ln|x+2| + C$

$= \frac{2}{9}\ln|x-1| - \frac{1}{3(x-1)} - \frac{2}{9}\ln|x+2| + C$

where $C$ is the constant of integration.

The logarithm terms can be combined using $\ln a - \ln b = \ln(a/b)$:

$= \frac{2}{9}(\ln|x-1| - \ln|x+2|) - \frac{1}{3(x-1)} + C$

$= \frac{2}{9}\ln\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C$

The final answer is $\textbf{$\frac{2}{9}\ln\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{3x + 5}{x^3 − x^2 − x + 1} dx$

First, we factor the denominator:

$x^3 − x^2 − x + 1 = x^2(x-1) - 1(x-1)$

$= (x^2 - 1)(x-1)$

$= (x-1)(x+1)(x-1)$

$= (x-1)^2(x+1)$

So, the integral is $\int \frac{3x + 5}{(x-1)^2(x+1)} dx$.

The denominator has a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x+1)$. We use partial fraction decomposition.

Let $\frac{3x + 5}{(x − 1)^2 (x + 1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$.

Multiply both sides by $(x-1)^2(x+1)$:

$3x + 5 = A(x-1)(x+1) + B(x+1) + C(x-1)^2$

$3x + 5 = A(x^2 - 1) + B(x+1) + C(x^2 - 2x + 1)$

To find A, B, and C, we substitute values for $x$.

Set $x = 1$:

$3(1) + 5 = A(1-1)(1+1) + B(1+1) + C(1-1)^2$

$8 = A(0)(2) + B(2) + C(0)^2$

$8 = 2B$

$B = 4$

Set $x = -1$:

$3(-1) + 5 = A(-1-1)(-1+1) + B(-1+1) + C(-1-1)^2$

$2 = A(-2)(0) + B(0) + C(-2)^2$

$2 = 4C$

$C = \frac{1}{2}$

To find A, compare coefficients or substitute another value for $x$, say $x=0$.

Set $x = 0$:

$3(0) + 5 = A(0-1)(0+1) + B(0+1) + C(0-1)^2$

$5 = A(-1)(1) + B(1) + C(-1)^2$

$5 = -A + B + C$

Substitute the values of B and C:

$5 = -A + 4 + \frac{1}{2}$

$5 = -A + \frac{8}{2} + \frac{1}{2}$

$5 = -A + \frac{9}{2}$

$A = \frac{9}{2} - 5 = \frac{9}{2} - \frac{10}{2} = -\frac{1}{2}$

$A = -\frac{1}{2}$

So, the partial fraction decomposition is:

$\frac{3x + 5}{(x − 1)^2 (x + 1)} = \frac{-1/2}{x-1} + \frac{4}{(x-1)^2} + \frac{1/2}{x+1}$

$= -\frac{1}{2(x-1)} + \frac{4}{(x-1)^2} + \frac{1}{2(x+1)}$

Now, we integrate the decomposed expression:

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} dx = \int \left(-\frac{1}{2(x-1)} + \frac{4}{(x-1)^2} + \frac{1}{2(x+1)}\right) dx$

$= -\frac{1}{2}\int \frac{1}{x-1} dx + 4\int (x-1)^{-2} dx + \frac{1}{2}\int \frac{1}{x+1} dx$

Integrating each term:

$\int \frac{1}{x-1} dx = \ln|x-1|$

$\int (x-1)^{-2} dx = \frac{(x-1)^{-1}}{-1} = -\frac{1}{x-1}$

$\int \frac{1}{x+1} dx = \ln|x+1|$

Combining the results and adding the constant of integration $C$:

$\int \frac{3x + 5}{x^3 − x^2 − x + 1} dx = -\frac{1}{2}\ln|x-1| + 4\left(-\frac{1}{x-1}\right) + \frac{1}{2}\ln|x+1| + C$

$= \frac{1}{2}\ln|x+1| - \frac{1}{2}\ln|x-1| - \frac{4}{x-1} + C$

Using logarithm properties, the answer can also be written as:

$\frac{1}{2}(\ln|x+1| - \ln|x-1|) - \frac{4}{x-1} + C = \frac{1}{2}\ln\left|\frac{x+1}{x-1}\right| - \frac{4}{x-1} + C$

The final answer is $\textbf{$\frac{1}{2}\ln\left|\frac{x+1}{x-1}\right| - \frac{4}{x-1} + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{2x - 3}{(x^2 - 1) (2x + 3)} dx$

First, we factor the denominator:

$x^2 - 1 = (x - 1)(x + 1)$

So the denominator is $(x - 1)(x + 1)(2x + 3)$. These are distinct linear factors.

We use the method of partial fraction decomposition.

Let $\frac{2x - 3}{(x - 1) (x + 1) (2x + 3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3}$.

Multiplying both sides by $(x-1)(x+1)(2x+3)$, we get:

$2x - 3 = A(x+1)(2x+3) + B(x-1)(2x+3) + C(x-1)(x+1)$

We find the values of A, B, and C by substituting the roots of the linear factors.

Setting $x = 1$:

$2(1) - 3 = A(1+1)(2(1)+3) + B(1-1)(2(1)+3) + C(1-1)(1+1)$

$2 - 3 = A(2)(5) + B(0)(5) + C(0)(2)$

$-1 = 10A$

$A = -\frac{1}{10}$

Setting $x = -1$:

$2(-1) - 3 = A(-1+1)(2(-1)+3) + B(-1-1)(2(-1)+3) + C(-1-1)(-1+1)$

$-2 - 3 = A(0)(1) + B(-2)(1) + C(-2)(0)$

$-5 = -2B$

$B = \frac{5}{2}$

Setting $x = -\frac{3}{2}$ (from $2x+3=0$):

$2\left(-\frac{3}{2}\right) - 3 = A\left(-\frac{3}{2}+1\right)\left(2\left(-\frac{3}{2}\right)+3\right) + B\left(-\frac{3}{2}-1\right)\left(2\left(-\frac{3}{2}\right)+3\right) + C\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}+1\right)$

$-3 - 3 = A\left(-\frac{1}{2}\right)(0) + B\left(-\frac{5}{2}\right)(0) + C\left(-\frac{5}{2}\right)\left(-\frac{1}{2}\right)$

$-6 = 0 + 0 + C\left(\frac{5}{4}\right)$

$-6 = \frac{5}{4}C$

$C = -6 \times \frac{4}{5} = -\frac{24}{5}$

$C = -\frac{24}{5}$

So, the partial fraction decomposition is:

$\frac{2x - 3}{(x - 1) (x + 1) (2x + 3)} = \frac{-1/10}{x-1} + \frac{5/2}{x+1} + \frac{-24/5}{2x+3}$

$= -\frac{1}{10(x-1)} + \frac{5}{2(x+1)} - \frac{24}{5(2x+3)}$

Now, we integrate the decomposed expression:

$\int \frac{2x - 3}{(x - 1) (x + 1) (2x + 3)} dx = \int \left(-\frac{1}{10(x-1)} + \frac{5}{2(x+1)} - \frac{24}{5(2x+3)}\right) dx$

$= -\frac{1}{10}\int \frac{1}{x-1} dx + \frac{5}{2}\int \frac{1}{x+1} dx - \frac{24}{5}\int \frac{1}{2x+3} dx$

Integrating each term:

$\int \frac{1}{x-1} dx = \ln|x-1|$

$\int \frac{1}{x+1} dx = \ln|x+1|$

For the third integral, let $u = 2x+3$, so $du = 2 dx$, which means $dx = \frac{1}{2} du$.

$\int \frac{1}{2x+3} dx = \int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{1}{2}\int \frac{1}{u} du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln|2x+3|$

Combining the results and adding the constant of integration $C$:

$\int \frac{2x - 3}{(x^2 - 1) (2x + 3)} dx = -\frac{1}{10}\ln|x-1| + \frac{5}{2}\ln|x+1| - \frac{24}{5}\left(\frac{1}{2}\ln|2x+3|\right) + C$

$= -\frac{1}{10}\ln|x-1| + \frac{5}{2}\ln|x+1| - \frac{12}{5}\ln|2x+3| + C$

The final answer is $\textbf{$-\frac{1}{10}\ln|x-1| + \frac{5}{2}\ln|x+1| - \frac{12}{5}\ln|2x+3| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{5x}{(x + 1) (x^2 − 4)} dx$

First, we factor the denominator completely:

$x^2 - 4 = (x - 2)(x + 2)$

So the denominator is $(x + 1)(x - 2)(x + 2)$, which are distinct linear factors.

We use the method of partial fraction decomposition.

Let $\frac{5x}{(x + 1) (x - 2) (x + 2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2}$.

Multiplying both sides by $(x+1)(x-2)(x+2)$, we get:

$5x = A(x-2)(x+2) + B(x+1)(x+2) + C(x+1)(x-2)$

We find the values of A, B, and C by substituting the roots of the linear factors.

Setting $x = -1$:

$5(-1) = A(-1-2)(-1+2) + B(-1+1)(-1+2) + C(-1+1)(-1-2)$

$-5 = A(-3)(1) + B(0)(1) + C(0)(-3)$

$-5 = -3A$

$A = \frac{5}{3}$

Setting $x = 2$:

$5(2) = A(2-2)(2+2) + B(2+1)(2+2) + C(2+1)(2-2)$

$10 = A(0)(4) + B(3)(4) + C(3)(0)$

$10 = 12B$

$B = \frac{10}{12} = \frac{5}{6}$

Setting $x = -2$:

$5(-2) = A(-2-2)(-2+2) + B(-2+1)(-2+2) + C(-2+1)(-2-2)$

$-10 = A(-4)(0) + B(-1)(0) + C(-1)(-4)$

$-10 = 4C$

$C = -\frac{10}{4} = -\frac{5}{2}$

So, the partial fraction decomposition is:

$\frac{5x}{(x + 1) (x^2 − 4)} = \frac{5/3}{x+1} + \frac{5/6}{x-2} + \frac{-5/2}{x+2}$

$= \frac{5}{3(x+1)} + \frac{5}{6(x-2)} - \frac{5}{2(x+2)}$

Now, we integrate the decomposed expression:

$\int \frac{5x}{(x + 1) (x^2 − 4)} dx = \int \left(\frac{5}{3(x+1)} + \frac{5}{6(x-2)} - \frac{5}{2(x+2)}\right) dx$

$= \frac{5}{3}\int \frac{1}{x+1} dx + \frac{5}{6}\int \frac{1}{x-2} dx - \frac{5}{2}\int \frac{1}{x+2} dx$

Integrating each term, we get:

$\int \frac{1}{x+1} dx = \ln|x+1|$

$\int \frac{1}{x-2} dx = \ln|x-2|$

$\int \frac{1}{x+2} dx = \ln|x+2|$

Combining the results and adding the constant of integration $C$:

$\int \frac{5x}{(x + 1) (x^2 − 4)} dx = \frac{5}{3}\ln|x+1| + \frac{5}{6}\ln|x-2| - \frac{5}{2}\ln|x+2| + C$

The final answer is $\textbf{$\frac{5}{3}\ln|x+1| + \frac{5}{6}\ln|x-2| - \frac{5}{2}\ln|x+2| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{x^3 + x + 1}{x^2 − 1} dx$

The degree of the numerator (3) is greater than the degree of the denominator (2). Therefore, it is an improper rational function, and we must perform polynomial long division first.

We divide $x^3 + x + 1$ by $x^2 - 1$.

$\begin{array}{r} x\phantom{+0} \\ x^2-1{\overline{\smash{\big)}\,x^3+0x^2+x+1\phantom{)}}}} \\ \underline{-~\phantom{(}(x^3-x)\phantom{+1)}} \\ 0+2x+1\phantom{)} \end{array}$

So, $\frac{x^3 + x + 1}{x^2 − 1} = x + \frac{2x + 1}{x^2 - 1}$.

The integral becomes:

$\int \left(x + \frac{2x + 1}{x^2 - 1}\right) dx = \int x dx + \int \frac{2x + 1}{x^2 - 1} dx$

Now, we integrate the second term $\int \frac{2x + 1}{x^2 - 1} dx$. We can use partial fraction decomposition for $\frac{2x + 1}{x^2 - 1}$.

Factor the denominator: $x^2 - 1 = (x-1)(x+1)$.

Let $\frac{2x + 1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$.

Multiplying by $(x-1)(x+1)$, we get:

$2x + 1 = A(x+1) + B(x-1)$

Set $x = 1$:

$2(1) + 1 = A(1+1) + B(1-1)$

$3 = 2A$

$A = \frac{3}{2}$

Set $x = -1$:

$2(-1) + 1 = A(-1+1) + B(-1-1)$

$-1 = -2B$

$B = \frac{1}{2}$

So, $\frac{2x + 1}{x^2 - 1} = \frac{3/2}{x-1} + \frac{1/2}{x+1}$.

Now, integrate each part of the original integral:

$\int \frac{x^3 + x + 1}{x^2 − 1} dx = \int x dx + \int \left(\frac{3/2}{x-1} + \frac{1/2}{x+1}\right) dx$

$= \int x dx + \frac{3}{2}\int \frac{1}{x-1} dx + \frac{1}{2}\int \frac{1}{x+1} dx$

Integrating each term:

$\int x dx = \frac{x^{1+1}}{1+1} + C_1 = \frac{x^2}{2} + C_1$

$\int \frac{1}{x-1} dx = \ln|x-1| + C_2$

$\int \frac{1}{x+1} dx = \ln|x+1| + C_3$

Combining the results and adding the constant of integration $C$:

$\int \frac{x^3 + x + 1}{x^2 − 1} dx = \frac{x^2}{2} + \frac{3}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + C$

The final answer is $\textbf{$\frac{x^2}{2} + \frac{3}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{2}{(1 − x) (1 + x^2)} dx$

The denominator has a linear factor $(1-x)$ and an irreducible quadratic factor $(1+x^2)$. We use the method of partial fraction decomposition.

Let $\frac{2}{(1 − x) (1 + x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2}$.

Multiplying both sides by $(1-x)(1+x^2)$, we get:

$2 = A(1+x^2) + (Bx+C)(1-x)$

Expanding the right side:

$2 = A + Ax^2 + Bx - Bx^2 + C - Cx$

Grouping terms by powers of $x$:

$2 = (A - B)x^2 + (B - C)x + (A + C)$

Comparing coefficients of corresponding powers of $x$ on both sides:

Coefficient of $x^2$: $A - B = 0$

Coefficient of $x$: $B - C = 0$

Constant term: $A + C = 2$

From the first two equations, we have $A = B$ and $B = C$. Thus, $A = B = C$.

Substitute $C = A$ into the third equation:

$A + A = 2$

$2A = 2$

$A = 1$

Since $A=B=C$, we have $A=1$, $B=1$, and $C=1$.

So, the partial fraction decomposition is:

$\frac{2}{(1 − x) (1 + x^2)} = \frac{1}{1-x} + \frac{1x+1}{1+x^2} = \frac{1}{1-x} + \frac{x+1}{1+x^2}$

Now, we integrate the decomposed expression:

$\int \frac{2}{(1 − x) (1 + x^2)} dx = \int \left(\frac{1}{1-x} + \frac{x+1}{1+x^2}\right) dx$

Split the integral into separate terms:

$= \int \frac{1}{1-x} dx + \int \frac{x+1}{1+x^2} dx$

$= \int \frac{1}{1-x} dx + \int \frac{x}{1+x^2} dx + \int \frac{1}{1+x^2} dx$

Integrate each term:

For the first integral, let $u = 1-x$, so $du = -dx$.

$\int \frac{1}{1-x} dx = \int \frac{1}{u} (-du) = -\int \frac{1}{u} du = -\ln|u| + C_1 = -\ln|1-x| + C_1$

For the second integral, let $v = 1+x^2$, so $dv = 2x dx$, which means $x dx = \frac{1}{2} dv$.

$\int \frac{x}{1+x^2} dx = \int \frac{1}{v} \left(\frac{1}{2} dv\right) = \frac{1}{2}\int \frac{1}{v} du = \frac{1}{2}\ln|v| + C_2 = \frac{1}{2}\ln(1+x^2) + C_2$ (since $1+x^2 > 0$)

For the third integral, it is a standard form:

$\int \frac{1}{1+x^2} dx = \arctan(x) + C_3$

Combining the results and the constants of integration into a single constant $C = C_1 + C_2 + C_3$:

$\int \frac{2}{(1 − x) (1 + x^2)} dx = -\ln|1-x| + \frac{1}{2}\ln(1+x^2) + \arctan(x) + C$

The final answer is $\textbf{$-\ln|1-x| + \frac{1}{2}\ln(1+x^2) + \arctan(x) + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{3x - 1}{(x + 2)^2} dx$

The denominator has a repeated linear factor $(x+2)^2$. We use the method of partial fraction decomposition.

Let $\frac{3x - 1}{(x + 2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}$.

Multiply both sides by $(x+2)^2$:

$3x - 1 = A(x+2) + B$

$3x - 1 = Ax + 2A + B$

Comparing coefficients of corresponding powers of $x$:

Coefficient of $x$: $3 = A$

Constant term: $-1 = 2A + B$

Substitute the value of $A$ into the second equation:

$-1 = 2(3) + B$

$-1 = 6 + B$

$B = -1 - 6$

$B = -7$

$A = 3$

So, the partial fraction decomposition is:

$\frac{3x - 1}{(x + 2)^2} = \frac{3}{x+2} + \frac{-7}{(x+2)^2} = \frac{3}{x+2} - \frac{7}{(x+2)^2}$

Now, we integrate the decomposed expression:

$\int \frac{3x - 1}{(x + 2)^2} dx = \int \left(\frac{3}{x+2} - \frac{7}{(x+2)^2}\right) dx$

$= \int \frac{3}{x+2} dx - \int \frac{7}{(x+2)^2} dx$

$= 3\int \frac{1}{x+2} dx - 7\int (x+2)^{-2} dx$

Integrating each term:

$\int \frac{1}{x+2} dx = \ln|x+2| + C_1$

$\int (x+2)^{-2} dx = \frac{(x+2)^{-2+1}}{-2+1} + C_2 = \frac{(x+2)^{-1}}{-1} + C_2 = -\frac{1}{x+2} + C_2$

Combining the results and adding the constant of integration $C$:

$\int \frac{3x - 1}{(x + 2)^2} dx = 3\ln|x+2| - 7\left(-\frac{1}{x+2}\right) + C$

$= 3\ln|x+2| + \frac{7}{x+2} + C$

where $C$ is the constant of integration.

The final answer is $\textbf{$3\ln|x+2| + \frac{7}{x+2} + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{1}{x^4 − 1} dx$

First, we factor the denominator completely:

$x^4 - 1 = (x^2 - 1)(x^2 + 1)$

$= (x - 1)(x + 1)(x^2 + 1)$

The denominator has two distinct linear factors $(x-1)$ and $(x+1)$ and an irreducible quadratic factor $(x^2+1)$. We use the method of partial fraction decomposition.

Let $\frac{1}{(x - 1)(x + 1)(x^2 + 1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$.

Multiply both sides by $(x-1)(x+1)(x^2+1)$:

$1 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+1)$

$1 = A(x^3 + x + x^2 + 1) + B(x^3 + x - x^2 - 1) + (Cx+D)(x^2 - 1)$

$1 = A(x^3 + x^2 + x + 1) + B(x^3 - x^2 + x - 1) + Cx^3 - Cx + Dx^2 - D$

Group terms by powers of $x$:

$1 = (A + B + C)x^3 + (A - B + D)x^2 + (A + B - C)x + (A - B - D)$

Comparing coefficients:

Coefficient of $x^3$: $A + B + C = 0$

Coefficient of $x^2$: $A - B + D = 0$

Coefficient of $x$: $A + B - C = 0$

Constant term: $A - B - D = 1$

From the first and third equations: $(A + B + C) - (A + B - C) = 0 - 0 \implies 2C = 0 \implies C = 0$.

Substituting $C=0$ into the first and third equations gives $A + B = 0$, so $B = -A$.

Substitute $B = -A$ into the second equation:

$A - (-A) + D = 0$

$2A + D = 0 \implies D = -2A$

Substitute $B = -A$ and $D = -2A$ into the fourth equation:

$A - (-A) - (-2A) = 1$

$A + A + 2A = 1$

$4A = 1$

$A = \frac{1}{4}$

Then, $B = -A = -\frac{1}{4}$.

And, $D = -2A = -2\left(\frac{1}{4}\right) = -\frac{1}{2}$.

So, the partial fraction decomposition is:

$\frac{1}{(x - 1)(x + 1)(x^2 + 1)} = \frac{1/4}{x-1} + \frac{-1/4}{x+1} + \frac{0x - 1/2}{x^2+1}$

$= \frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)}$

Now, we integrate the decomposed expression:

$\int \frac{1}{x^4 − 1} dx = \int \left(\frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)}\right) dx$

$= \frac{1}{4}\int \frac{1}{x-1} dx - \frac{1}{4}\int \frac{1}{x+1} dx - \frac{1}{2}\int \frac{1}{x^2+1} dx$

Integrating each term:

$\int \frac{1}{x-1} dx = \ln|x-1|$

$\int \frac{1}{x+1} dx = \ln|x+1|$

$\int \frac{1}{x^2+1} dx = \arctan(x)$

Combining the results and adding the constant of integration $C$:

$\int \frac{1}{x^4 − 1} dx = \frac{1}{4}\ln|x-1| - \frac{1}{4}\ln|x+1| - \frac{1}{2}\arctan(x) + C$

Using logarithm properties, the answer can also be written as:

$\frac{1}{4}(\ln|x-1| - \ln|x+1|) - \frac{1}{2}\arctan(x) + C = \frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\arctan(x) + C$

The final answer is $\textbf{$\frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\arctan(x) + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral:

$\int \frac{1}{x(x^n + 1)} dx$

Following the hint, we multiply the numerator and the denominator by $x^{n-1}$:

$\int \frac{1 \cdot x^{n-1}}{x(x^n + 1) \cdot x^{n-1}} dx = \int \frac{x^{n-1}}{x^n(x^n + 1)} dx$

Now, we use the substitution $t = x^n$.

Differentiating with respect to $x$, we get $\frac{dt}{dx} = nx^{n-1}$.

This implies $dt = nx^{n-1} dx$, so $x^{n-1} dx = \frac{1}{n} dt$.

Substitute $t$ and $dx$ into the integral:

$\int \frac{x^{n-1}}{x^n(x^n + 1)} dx = \int \frac{1}{t(t + 1)} \left(\frac{1}{n} dt\right) = \frac{1}{n} \int \frac{1}{t(t + 1)} dt$

Now we need to evaluate the integral $\int \frac{1}{t(t+1)} dt$. We use partial fraction decomposition for $\frac{1}{t(t+1)}$.

Let $\frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1}$.

Multiplying both sides by $t(t+1)$, we get:

$1 = A(t+1) + Bt$

Setting $t = 0$:

$1 = A(0+1) + B(0)$

$1 = A$

Setting $t = -1$:

$1 = A(-1+1) + B(-1)$

$1 = -B$

$B = -1$

So, the partial fraction decomposition is:

$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$

Now integrate with respect to $t$:

$\int \frac{1}{t(t + 1)} dt = \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt$

$= \int \frac{1}{t} dt - \int \frac{1}{t+1} dt$

$= \ln|t| - \ln|t+1| + C'$

Using logarithm properties, this can be written as $\ln\left|\frac{t}{t+1}\right| + C'$.

Substitute this back into the main integral expression:

$\int \frac{1}{x(x^n + 1)} dx = \frac{1}{n} \int \frac{1}{t(t + 1)} dt = \frac{1}{n} \left(\ln\left|\frac{t}{t+1}\right| + C'\right)$

$= \frac{1}{n}\ln\left|\frac{t}{t+1}\right| + \frac{C'}{n}$

Finally, substitute back $t = x^n$ and let $C = \frac{C'}{n}$ be the constant of integration:

$\int \frac{1}{x(x^n + 1)} dx = \frac{1}{n}\ln\left|\frac{x^n}{x^n+1}\right| + C$

The final answer is $\textbf{$\frac{1}{n}\ln\left|\frac{x^n}{x^n+1}\right| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral:

$\int \frac{\cos x}{(1 − \sin x) (2 − \sin x)} dx$

Following the hint, we use the substitution $t = \sin x$.

Differentiating with respect to $x$, we get $\frac{dt}{dx} = \cos x$.

This implies $dt = \cos x dx$.

Substitute $t$ and $dx$ into the integral:

$\int \frac{\cos x}{(1 − \sin x) (2 − \sin x)} dx = \int \frac{1}{(1 - t)(2 - t)} dt$

Now we need to evaluate the integral $\int \frac{1}{(1 - t)(2 - t)} dt$. We use partial fraction decomposition.

Let $\frac{1}{(1 - t)(2 - t)} = \frac{A}{1-t} + \frac{B}{2-t}$.

Multiply both sides by $(1-t)(2-t)$:

$1 = A(2-t) + B(1-t)$

Set $t = 1$:

$1 = A(2-1) + B(1-1)$

$1 = A(1) + B(0)$

$A = 1$

Set $t = 2$:

$1 = A(2-2) + B(1-2)$

$1 = A(0) + B(-1)$

$1 = -B$

$B = -1$

So, the partial fraction decomposition is:

$\frac{1}{(1 - t)(2 - t)} = \frac{1}{1-t} + \frac{-1}{2-t} = \frac{1}{1-t} - \frac{1}{2-t}$

Now integrate with respect to $t$:

$\int \frac{1}{(1 - t)(2 - t)} dt = \int \left(\frac{1}{1-t} - \frac{1}{2-t}\right) dt$

$= \int \frac{1}{1-t} dt - \int \frac{1}{2-t} dt$

For the first integral, let $u = 1-t$, $du = -dt$. $\int \frac{1}{u}(-du) = -\ln|u| = -\ln|1-t|$.

For the second integral, let $v = 2-t$, $dv = -dt$. $\int \frac{1}{v}(-dv) = -\ln|v| = -\ln|2-t|$.

So, the integral is:

$(-\ln|1-t|) - (-\ln|2-t|) + C'$

$= -\ln|1-t| + \ln|2-t| + C'$

$= \ln|2-t| - \ln|1-t| + C'$

Using logarithm properties, this is $\ln\left|\frac{2-t}{1-t}\right| + C'$.

Finally, substitute back $t = \sin x$ and let $C = C'$ be the constant of integration:

$\int \frac{\cos x}{(1 − \sin x) (2 − \sin x)} dx = \ln\left|\frac{2-\sin x}{1-\sin x}\right| + C$

The final answer is $\textbf{$\ln\left|\frac{2-\sin x}{1-\sin x}\right| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} dx$

First, simplify the numerator and denominator:

Numerator: $(x^2 + 1)(x^2 + 2) = x^4 + 2x^2 + x^2 + 2 = x^4 + 3x^2 + 2$

Denominator: $(x^2 + 3)(x^2 + 4) = x^4 + 4x^2 + 3x^2 + 12 = x^4 + 7x^2 + 12$

The rational function is $\frac{x^4 + 3x^2 + 2}{x^4 + 7x^2 + 12}$. The degree of the numerator (4) is equal to the degree of the denominator (4). Therefore, it is an improper rational function, and we must perform polynomial long division first.

We divide $x^4 + 3x^2 + 2$ by $x^4 + 7x^2 + 12$.

$\begin{array}{r} 1\phantom{x^4+7x^2+12)} \\ x^4+7x^2+12{\overline{\smash{\big)}\,x^4+3x^2+2\phantom{)}}}} \\ \underline{-~\phantom{(}(x^4+7x^2+12)} \\ 0-4x^2-10\phantom{)} \end{array}$

So, $\frac{x^4 + 3x^2 + 2}{x^4 + 7x^2 + 12} = 1 + \frac{-4x^2 - 10}{x^4 + 7x^2 + 12} = 1 - \frac{4x^2 + 10}{x^4 + 7x^2 + 12}$.

Substitute the factored denominator back:

$= 1 - \frac{4x^2 + 10}{(x^2 + 3) (x^2 + 4)}$

The integral becomes:

$\int \left(1 - \frac{4x^2 + 10}{(x^2 + 3) (x^2 + 4)}\right) dx = \int 1 dx - \int \frac{4x^2 + 10}{(x^2 + 3) (x^2 + 4)} dx$

Now, we use partial fraction decomposition for $\frac{4x^2 + 10}{(x^2 + 3) (x^2 + 4)}$.

Since the denominator has irreducible quadratic factors, the numerator in the partial fraction should be linear.

Let $\frac{4x^2 + 10}{(x^2 + 3) (x^2 + 4)} = \frac{Ax+B}{x^2+3} + \frac{Cx+D}{x^2+4}$.

Alternatively, since the expression only contains $x^2$ terms, we can substitute $y = x^2$ temporarily to find the coefficients for $\frac{4y + 10}{(y + 3) (y + 4)}$.

Let $\frac{4y + 10}{(y + 3) (y + 4)} = \frac{A'}{y+3} + \frac{B'}{y+4}$.

$4y + 10 = A'(y+4) + B'(y+3)$

Set $y = -3$:

$4(-3) + 10 = A'(-3+4) + B'(-3+3)$

$-12 + 10 = A'(1) + B'(0)$

$-2 = A'$

Set $y = -4$:

$4(-4) + 10 = A'(-4+4) + B'(-4+3)$

$-16 + 10 = A'(0) + B'(-1)$

$-6 = -B'$

$B' = 6$

So, $\frac{4y + 10}{(y + 3) (y + 4)} = \frac{-2}{y+3} + \frac{6}{y+4}$.

Substituting back $y = x^2$, we get:

$\frac{4x^2 + 10}{(x^2 + 3) (x^2 + 4)} = \frac{-2}{x^2+3} + \frac{6}{x^2+4}$

Now the partial fraction form is simpler, where $A=0$, $B=-2$, $C=0$, $D=6$ in the $\frac{Ax+B}{x^2+3} + \frac{Cx+D}{x^2+4}$ form.

Now integrate the decomposed terms:

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} dx = \int 1 dx - \int \left(\frac{-2}{x^2+3} + \frac{6}{x^2+4}\right) dx$

$= \int 1 dx + 2\int \frac{1}{x^2+(\sqrt{3})^2} dx - 6\int \frac{1}{x^2+2^2} dx$

Integrating each term:

$\int 1 dx = x + C_1$

$\int \frac{1}{x^2+a^2} dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)$.

$2\int \frac{1}{x^2+(\sqrt{3})^2} dx = 2 \cdot \frac{1}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) + C_2 = \frac{2}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) + C_2$

$6\int \frac{1}{x^2+2^2} dx = 6 \cdot \frac{1}{2}\arctan\left(\frac{x}{2}\right) + C_3 = 3\arctan\left(\frac{x}{2}\right) + C_3$

Combining the results and adding the constant of integration $C$:

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} dx = x + \frac{2}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) - 3\arctan\left(\frac{x}{2}\right) + C$

The final answer is $\textbf{$x + \frac{2}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) - 3\arctan\left(\frac{x}{2}\right) + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{2x}{(x^2 + 1) (x^2 + 3)} dx$

We can use substitution. Let $t = x^2$.

Differentiating with respect to $x$, we get $\frac{dt}{dx} = 2x$.

This implies $dt = 2x dx$.

Substitute $t$ and $dx$ into the integral:

$\int \frac{2x}{(x^2 + 1) (x^2 + 3)} dx = \int \frac{1}{(t + 1) (t + 3)} dt$

Now we need to evaluate the integral $\int \frac{1}{(t + 1) (t + 3)} dt$. We use partial fraction decomposition.

Let $\frac{1}{(t + 1) (t + 3)} = \frac{A}{t+1} + \frac{B}{t+3}$.

Multiply both sides by $(t+1)(t+3)$:

$1 = A(t+3) + B(t+1)$

Set $t = -1$:

$1 = A(-1+3) + B(-1+1)$

$1 = A(2) + B(0)$

$2A = 1 \implies A = \frac{1}{2}$

Set $t = -3$:

$1 = A(-3+3) + B(-3+1)$

$1 = A(0) + B(-2)$

$-2B = 1 \implies B = -\frac{1}{2}$

So, the partial fraction decomposition is:

$\frac{1}{(t + 1) (t + 3)} = \frac{1/2}{t+1} + \frac{-1/2}{t+3} = \frac{1}{2(t+1)} - \frac{1}{2(t+3)}$

Now integrate with respect to $t$:

$\int \frac{1}{(t + 1) (t + 3)} dt = \int \left(\frac{1}{2(t+1)} - \frac{1}{2(t+3)}\right) dt$

$= \frac{1}{2}\int \frac{1}{t+1} dt - \frac{1}{2}\int \frac{1}{t+3} dt$

$= \frac{1}{2}\ln|t+1| - \frac{1}{2}\ln|t+3| + C'$

Using logarithm properties, this is $\frac{1}{2}(\ln|t+1| - \ln|t+3|) + C' = \frac{1}{2}\ln\left|\frac{t+1}{t+3}\right| + C'$.

Finally, substitute back $t = x^2$ and let $C = C'$ be the constant of integration:

$\int \frac{2x}{(x^2 + 1) (x^2 + 3)} dx = \frac{1}{2}\ln\left|\frac{x^2+1}{x^2+3}\right| + C$

Since $x^2+1 > 0$ and $x^2+3 > 0$, the absolute values can be removed.

The final answer is $\textbf{$\frac{1}{2}\ln\left(\frac{x^2+1}{x^2+3}\right) + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral of the given rational function:

$\int \frac{1}{x(x^4 − 1)} dx$

We can rewrite the integral by multiplying the numerator and denominator by $x^3$:

$\int \frac{1 \cdot x^3}{x(x^4 − 1) \cdot x^3} dx = \int \frac{x^3}{x^4(x^4 − 1)} dx$

Now, we use the substitution method. Let $t = x^4$.

Differentiating with respect to $x$, we get $\frac{dt}{dx} = 4x^3$.

This means $dt = 4x^3 dx$, or $x^3 dx = \frac{1}{4} dt$.

Substitute $t$ and $dx$ into the integral:

$\int \frac{x^3}{x^4(x^4 − 1)} dx = \int \frac{1}{t(t - 1)} \left(\frac{1}{4} dt\right) = \frac{1}{4} \int \frac{1}{t(t - 1)} dt$

Now we need to evaluate the integral $\int \frac{1}{t(t - 1)} dt$. We use partial fraction decomposition for $\frac{1}{t(t - 1)}$.

Let $\frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t-1}$.

Multiply both sides by $t(t-1)$:

$1 = A(t-1) + Bt$

Setting $t = 0$:

$1 = A(0-1) + B(0)$

$1 = -A \implies A = -1$

Setting $t = 1$:

$1 = A(1-1) + B(1)$

$1 = 0 + B \implies B = 1$

So, the partial fraction decomposition is:

$\frac{1}{t(t - 1)} = \frac{-1}{t} + \frac{1}{t-1} = \frac{1}{t-1} - \frac{1}{t}$

Now integrate with respect to $t$:

$\int \frac{1}{t(t - 1)} dt = \int \left(\frac{1}{t-1} - \frac{1}{t}\right) dt$

$= \int \frac{1}{t-1} dt - \int \frac{1}{t} dt$

$= \ln|t-1| - \ln|t| + C'$

Using logarithm properties, this is $\ln\left|\frac{t-1}{t}\right| + C'$.

Substitute this back into the main integral expression:

$\int \frac{1}{x(x^4 - 1)} dx = \frac{1}{4} \int \frac{1}{t(t - 1)} dt = \frac{1}{4} \left(\ln\left|\frac{t-1}{t}\right| + C'\right)$

$= \frac{1}{4}\ln\left|\frac{t-1}{t}\right| + \frac{C'}{4}$

Finally, substitute back $t = x^4$ and let $C = \frac{C'}{4}$ be the constant of integration:

$\int \frac{1}{x(x^4 − 1)} dx = \frac{1}{4}\ln\left|\frac{x^4-1}{x^4}\right| + C$

The final answer is $\textbf{$\frac{1}{4}\ln\left|\frac{x^4-1}{x^4}\right| + C$}$

or equivalently, $\textbf{$\frac{1}{4}\ln\left|1 - \frac{1}{x^4}\right| + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral:

$\int \frac{1}{e^x - 1} dx$

Following the hint, we use the substitution $t = e^x$.

Differentiating with respect to $x$, we get $\frac{dt}{dx} = e^x$.

This means $dt = e^x dx$. Since $t = e^x$, we have $dx = \frac{dt}{e^x} = \frac{dt}{t}$.

Substitute $t$ and $dx$ into the integral:

$\int \frac{1}{e^x - 1} dx = \int \frac{1}{t - 1} \left(\frac{dt}{t}\right) = \int \frac{1}{t(t - 1)} dt$

Now we need to evaluate the integral $\int \frac{1}{t(t - 1)} dt$. We use partial fraction decomposition.

Let $\frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t-1}$.

Multiply both sides by $t(t-1)$:

$1 = A(t-1) + Bt$

Setting $t = 0$:

$1 = A(0-1) + B(0)$

$1 = -A \implies A = -1$

Setting $t = 1$:

$1 = A(1-1) + B(1)$

$1 = 0 + B \implies B = 1$

So, the partial fraction decomposition is:

$\frac{1}{t(t - 1)} = \frac{-1}{t} + \frac{1}{t-1} = \frac{1}{t-1} - \frac{1}{t}$

Now integrate with respect to $t$:

$\int \frac{1}{t(t - 1)} dt = \int \left(\frac{1}{t-1} - \frac{1}{t}\right) dt$

$= \int \frac{1}{t-1} dt - \int \frac{1}{t} dt$

$= \ln|t-1| - \ln|t| + C'$

Using logarithm properties, this is $\ln\left|\frac{t-1}{t}\right| + C'$.

Finally, substitute back $t = e^x$ and let $C = C'$ be the constant of integration:

$\int \frac{1}{e^x - 1} dx = \ln\left|\frac{e^x-1}{e^x}\right| + C$

The term inside the logarithm can be written as $\frac{e^x-1}{e^x} = 1 - \frac{1}{e^x} = 1 - e^{-x}$.

So the result is $\ln|1 - e^{-x}| + C$.

Alternatively, using $\ln|t-1| - \ln|t| = \ln|e^x - 1| - \ln|e^x| = \ln|e^x - 1| - x$ (since $\ln|e^x| = x$ for real $x$).

The final answer is $\textbf{$\ln\left|\frac{e^x-1}{e^x}\right| + C$}$

or

$\textbf{$\ln|e^x - 1| - x + C$}$

where $C$ is the constant of integration.

We need to evaluate the integral:

$\int \frac{x}{(x - 1) (x - 2)} dx$

We use the method of partial fraction decomposition for the integrand $\frac{x}{(x - 1) (x - 2)}$.

Let $\frac{x}{(x - 1) (x - 2)} = \frac{A}{x-1} + \frac{B}{x-2}$.

Multiplying both sides by $(x-1)(x-2)$, we get:

$x = A(x-2) + B(x-1)$

To find the coefficients $A$ and $B$, we substitute the roots of the denominator.

Setting $x = 1$:

$1 = A(1-2) + B(1-1)$

$1 = A(-1) + B(0)$

$1 = -A \implies A = -1$

Setting $x = 2$:

$2 = A(2-2) + B(2-1)$

$2 = A(0) + B(1)$

$2 = B$

So, the partial fraction decomposition is:

$\frac{x}{(x - 1) (x - 2)} = \frac{-1}{x-1} + \frac{2}{x-2}$

Now, we integrate the decomposed expression:

$\int \frac{x}{(x - 1) (x - 2)} dx = \int \left(\frac{-1}{x-1} + \frac{2}{x-2}\right) dx$

$= \int \frac{-1}{x-1} dx + \int \frac{2}{x-2} dx$

$= -\int \frac{1}{x-1} dx + 2\int \frac{1}{x-2} dx$

Integrating each term, we get:

$-\int \frac{1}{x-1} dx = -\ln|x-1| + C_1$

$2\int \frac{1}{x-2} dx = 2\ln|x-2| + C_2$

Combining the results and the constants of integration into a single constant $C = C_1 + C_2$:

$\int \frac{x}{(x - 1) (x - 2)} dx = -\ln|x-1| + 2\ln|x-2| + C$

Using logarithm properties ($m\ln a = \ln a^m$ and $\ln a - \ln b = \ln \frac{a}{b}$), we can rewrite the expression:

$2\ln|x-2| - \ln|x-1| + C = \ln|(x-2)^2| - \ln|x-1| + C = \ln\left|\frac{(x-2)^2}{x-1}\right| + C$

Comparing this result with the given options:

(A) $\log \left| \frac{(x − 1)^2}{x − 2} \right| + C = \ln\left|\frac{(x-1)^2}{x-2}\right| + C$ (using natural logarithm $\log$) = $2\ln|x-1| - \ln|x-2| + C$. This does not match.

(B) $\log \left| \frac{(x − 2)^2}{x − 1} \right| + C = \ln\left|\frac{(x-2)^2}{x-1}\right| + C$. This matches our result.

(C) $\log \left| \left( \frac{x − 1}{x − 2} \right)^2 \right| + C = 2\log \left| \frac{x − 1}{x − 2} \right| + C = 2(\ln|x-1| - \ln|x-2|) + C = 2\ln|x-1| - 2\ln|x-2| + C$. This does not match.

(D) log |(x - 1) (x - 2)| + c = $\ln|(x-1)(x-2)| + C = \ln|x-1| + \ln|x-2| + C$. This does not match.

The correct option is (B).

The final answer is (B).

We need to evaluate the integral:

$\int \frac{1}{x (x^2 + 1)} dx$

We use the method of partial fraction decomposition for the integrand $\frac{1}{x(x^2 + 1)}$. The denominator has a linear factor $x$ and an irreducible quadratic factor $x^2+1$.

Let $\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$.

Multiplying both sides by $x(x^2+1)$, we get:

$1 = A(x^2+1) + (Bx+C)x$

$1 = Ax^2 + A + Bx^2 + Cx$

$1 = (A+B)x^2 + Cx + A$

Comparing coefficients of corresponding powers of $x$:

Coefficient of $x^2$: $A+B = 0$

Coefficient of $x$: $C = 0$

Constant term: $A = 1$

From $A=1$ and $A+B=0$, we have $1+B=0$, which gives $B=-1$. From the coefficient of $x$, we have $C=0$.

So, the partial fraction decomposition is:

$\frac{1}{x(x^2 + 1)} = \frac{1}{x} + \frac{-1x+0}{x^2+1} = \frac{1}{x} - \frac{x}{x^2+1}$

Now, we integrate the decomposed expression:

$\int \frac{1}{x(x^2 + 1)} dx = \int \left(\frac{1}{x} - \frac{x}{x^2+1}\right) dx$

$= \int \frac{1}{x} dx - \int \frac{x}{x^2+1} dx$

The first integral is standard:

$\int \frac{1}{x} dx = \ln|x| + C_1$

For the second integral, $\int \frac{x}{x^2+1} dx$, we use substitution. Let $u = x^2+1$. Then $du = 2x dx$, so $x dx = \frac{1}{2} du$.

$\int \frac{x}{x^2+1} dx = \int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2}\ln|u| + C_2$

Substituting back $u = x^2+1$, we get $\frac{1}{2}\ln|x^2+1| + C_2$. Since $x^2+1 > 0$ for all real $x$, we can write this as $\frac{1}{2}\ln(x^2+1) + C_2$.

Combining the results and the constants of integration into a single constant $C = C_1 - C_2$:

$\int \frac{1}{x (x^2 + 1)} dx = \ln|x| - \frac{1}{2}\ln(x^2+1) + C$

We compare this result with the given options. Note that in multiple choice questions involving logarithms, $\log$ usually refers to the natural logarithm $\ln$.

(A) $\log |x| - \frac{1}{2} \log (x^2 + 1) + C$. This matches our result $\ln|x| - \frac{1}{2}\ln(x^2+1) + C$.

(B) $\log |x| + \frac{1}{2} \log (x^2 + 1) + C$. The sign of the second term is incorrect.

(C) $-\log |x| + \frac{1}{2} \log (x^2 + 1) + C$. The sign of the first term is incorrect.

(D) $\frac{1}{2} \log |x| + \log (x^2 + 1) + C$. The coefficients are incorrect.

The correct option is (A).

The final answer is (A).

We need to evaluate the integral:

$\int x \cos x \;dx$

We use the method of integration by parts. The formula for integration by parts is:

$\int u \; dv = uv - \int v \; du$

We choose $u$ and $dv$ from the integrand $x \cos x \;dx$. Using the LIATE rule, we choose $u = x$ (Algebraic) and $dv = \cos x \;dx$ (Trigonometric).

Determine $du$ by differentiating $u$ and $v$ by integrating $dv$:

$u = x \implies du = dx$

$dv = \cos x \;dx \implies v = \int \cos x \;dx = \sin x$

Apply the integration by parts formula $\int u \; dv = uv - \int v \; du$:

$\int x \cos x \;dx = x (\sin x) - \int (\sin x) \; dx$

$\int x \cos x \;dx = x \sin x - \int \sin x \; dx$

Now, evaluate the remaining integral $\int \sin x \; dx$:

$\int \sin x \; dx = -\cos x + C'$

Substitute this result back into the equation:

$\int x \cos x \;dx = x \sin x - (-\cos x) + C$

$\int x \cos x \;dx = x \sin x + \cos x + C$

where $C$ is the constant of integration.

The final answer is $\textbf{$x \sin x + \cos x + C$}$.

where $C$ is the constant of integration.

We need to evaluate the integral:

$\int \log x \;dx$

We assume that $\log x$ refers to the natural logarithm, $\ln x$. The integral is $\int \ln x \;dx$.

We use the method of integration by parts. The formula is $\int u \; dv = uv - \int v \; du$.

To apply integration by parts, we need to express the integrand as a product of two functions. We can write $\ln x$ as $\ln x \cdot 1$.

Using the LIATE rule, we choose $u = \ln x$ (Logarithmic) and $dv = 1 \;dx$ (Algebraic).

Determine $du$ by differentiating $u$ and $v$ by integrating $dv$:

$u = \ln x \implies du = \frac{1}{x} dx$

$dv = 1 \;dx \implies v = \int 1 \;dx = x$

Apply the integration by parts formula $\int u \; dv = uv - \int v \; du$:

$\int \ln x \;dx = (\ln x) (x) - \int x \left(\frac{1}{x} dx\right)$

$\int \ln x \;dx = x \ln x - \int 1 \; dx$

Now, evaluate the remaining integral $\int 1 \; dx$:

$\int 1 \; dx = x + C'$

Substitute this result back into the equation:

$\int \ln x \;dx = x \ln x - (x) + C$

$\int \ln x \;dx = x \ln x - x + C$

where $C$ is the constant of integration.

The result can also be written as $x(\ln x - 1) + C$.

The final answer is $\textbf{$x \ln x - x + C$}$.

where $C$ is the constant of integration.

We need to evaluate the integral:

$\int x \ e^x \;dx$

We use the method of integration by parts. The formula is $\int u \; dv = uv - \int v \; du$.

Using the LIATE rule, we choose $u = x$ (Algebraic) and $dv = e^x \;dx$ (Exponential).

Determine $du$ by differentiating $u$ and $v$ by integrating $dv$:

$u = x \implies du = dx$

$dv = e^x \;dx \implies v = \int e^x \;dx = e^x$

Apply the integration by parts formula $\int u \; dv = uv - \int v \; du$:

$\int x \ e^x \;dx = (x) (e^x) - \int e^x \; dx$

$\int x \ e^x \;dx = x e^x - \int e^x \; dx$

Now, evaluate the remaining integral $\int e^x \; dx$:

$\int e^x \; dx = e^x + C'$

Substitute this result back into the equation:

$\int x \ e^x \;dx = x e^x - (e^x) + C$

$\int x \ e^x \;dx = x e^x - e^x + C$

where $C$ is the constant of integration.

The result can also be written as $e^x(x - 1) + C$.

The final answer is $\textbf{$x e^x - e^x + C$}$.

where $C$ is the constant of integration.

We need to evaluate the integral:

$\int \frac{x \sin^{−1} x}{\sqrt{1 − x^2}} \;dx$

We use the substitution method. Let $t = \sin^{-1} x$.

Differentiating with respect to $x$, we get:

$\frac{dt}{dx} = \frac{1}{\sqrt{1-x^2}}$

So, $dt = \frac{1}{\sqrt{1-x^2}} dx$.

Also, from $t = \sin^{-1} x$, we have $x = \sin t$.

Substitute $t$, $x$, and $dx$ in terms of $t$ and $dt$ into the integral:

$\int \frac{x \sin^{−1} x}{\sqrt{1 − x^2}} \;dx = \int \sin t \cdot t \cdot \frac{1}{\sqrt{1-x^2}} dx = \int t \sin t \; dt$

Now we evaluate $\int t \sin t \; dt$ using integration by parts. The formula is $\int u \; dv = uv - \int v \; du$.

Using the LIATE rule, we choose $u = t$ (Algebraic) and $dv = \sin t \; dt$ (Trigonometric).

Determine $du$ by differentiating $u$ and $v$ by integrating $dv$:

$u = t \implies du = dt$

$dv = \sin t \; dt \implies v = \int \sin t \; dt = -\cos t$

Apply the integration by parts formula $\int u \; dv = uv - \int v \; du$:

$\int t \sin t \; dt = (t)(-\cos t) - \int (-\cos t) \; dt$

$= -t \cos t + \int \cos t \; dt$

Evaluate the remaining integral $\int \cos t \; dt$:

$\int \cos t \; dt = \sin t + C'$

Substitute this result back into the integration by parts formula:

$\int t \sin t \; dt = -t \cos t + \sin t + C'$

Finally, substitute back $t = \sin^{-1} x$. We need to express $\cos t$ in terms of $x$.

Since $t = \sin^{-1} x$, we have $\sin t = x$. Using the identity $\sin^2 t + \cos^2 t = 1$, we get $\cos^2 t = 1 - \sin^2 t = 1 - x^2$. So, $\cos t = \pm \sqrt{1 - x^2}$. Assuming the principal value of $\sin^{-1}x$, $t \in [-\pi/2, \pi/2]$, where $\cos t \ge 0$. Thus, $\cos t = \sqrt{1 - x^2}$.

Substitute $t = \sin^{-1} x$, $\cos t = \sqrt{1 - x^2}$, and $\sin t = x$ back into the expression in terms of $t$:

$\int \frac{x \sin^{−1} x}{\sqrt{1 − x^2}} \;dx = -(\sin^{-1} x) (\sqrt{1 - x^2}) + x + C$

$= x - \sqrt{1 - x^2} \sin^{-1} x + C$

where $C$ is the constant of integration.

The final answer is $\textbf{$x - \sqrt{1 - x^2} \sin^{-1} x + C$}$.

where $C$ is the constant of integration.

We need to evaluate the integral:

$I = \int e^x \sin x \;dx$

We use the method of integration by parts, which has the formula $\int u \; dv = uv - \int v \; du$.

Let $u = \sin x$ and $dv = e^x \;dx$.

Determine $du$ by differentiating $u$ and $v$ by integrating $dv$:

$u = \sin x \implies du = \cos x \;dx$

$dv = e^x \;dx \implies v = \int e^x \;dx = e^x$

Apply the integration by parts formula:

$I = (\sin x)(e^x) - \int e^x (\cos x \;dx)$

$I = e^x \sin x - \int e^x \cos x \;dx$

The new integral $\int e^x \cos x \;dx$ also requires integration by parts. Let's evaluate this integral separately or continue by applying integration by parts to this term.

For the integral $\int e^x \cos x \;dx$, let $u' = \cos x$ and $dv' = e^x \;dx$.

Determine $du'$ and $v'$:

$u' = \cos x \implies du' = -\sin x \;dx$

$dv' = e^x \;dx \implies v' = \int e^x \;dx = e^x$

Apply the integration by parts formula to $\int e^x \cos x \;dx$:

$\int e^x \cos x \;dx = (\cos x)(e^x) - \int e^x (-\sin x \;dx)$

$\int e^x \cos x \;dx = e^x \cos x + \int e^x \sin x \;dx$

Substitute this result back into the equation for $I$:

$I = e^x \sin x - (e^x \cos x + \int e^x \sin x \;dx)$

$I = e^x \sin x - e^x \cos x - \int e^x \sin x \;dx$

Notice that the original integral $I = \int e^x \sin x \;dx$ appears on the right side. Let's move it to the left side:

$I + I = e^x \sin x - e^x \cos x$

$2I = e^x \sin x - e^x \cos x$

Divide by 2 to solve for $I$, and add the constant of integration $C$:

$I = \frac{1}{2}(e^x \sin x - e^x \cos x) + C$

The final answer is $\textbf{$\frac{1}{2}e^x (\sin x - \cos x) + C$}$.

where $C$ is the constant of integration.

We need to evaluate the given integrals.

(i) Find $\int e^x \left( \tan^{−1} x + \frac{1}{1 + x^2} \right) \;dx$

This integral is of the form $\int e^x [f(x) + f'(x)] dx$.

Recall the formula: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.

Let $f(x) = \tan^{-1} x$.

Differentiating $f(x)$ with respect to $x$, we get:

$f'(x) = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$

The integrand is $e^x [f(x) + f'(x)]$.

Using the formula, the integral is:

$\int e^x \left( \tan^{−1} x + \frac{1}{1 + x^2} \right) \;dx = e^x \tan^{-1} x + C$

where $C$ is the constant of integration.

The answer for (i) is $\textbf{$e^x \tan^{-1} x + C$}$.

(ii) Find $\int \frac{(x^2 + 1) e^x}{(x + 1)^2} \;dx$

We aim to rewrite the integrand in the form $e^x [f(x) + f'(x)]$.

Let's manipulate the term $\frac{x^2 + 1}{(x + 1)^2}$.

We can add and subtract 1 from the numerator:

$\frac{x^2 + 1}{(x + 1)^2} = \frac{x^2 - 1 + 2}{(x + 1)^2} = \frac{(x-1)(x+1) + 2}{(x + 1)^2}$

$= \frac{(x-1)(x+1)}{(x+1)^2} + \frac{2}{(x + 1)^2} = \frac{x-1}{x+1} + \frac{2}{(x+1)^2}$

So, the integrand is $e^x \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^2} \right)$.

Let $f(x) = \frac{x-1}{x+1}$. We need to find its derivative $f'(x)$.

Using the quotient rule, $f'(x) = \frac{d}{dx}\left(\frac{x-1}{x+1}\right) = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2}$

$= \frac{x+1 - (x-1)}{(x+1)^2} = \frac{x+1 - x + 1}{(x+1)^2} = \frac{2}{(x+1)^2}$

We can see that $f(x) = \frac{x-1}{x+1}$ and $f'(x) = \frac{2}{(x+1)^2}$.

The integrand is in the form $e^x [f(x) + f'(x)]$.

Using the formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$, the integral is:

$\int \frac{(x^2 + 1) e^x}{(x + 1)^2} \;dx = \int e^x \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^2} \right) \;dx = e^x \left(\frac{x-1}{x+1}\right) + C$

where $C$ is the constant of integration.

The answer for (ii) is $\textbf{$e^x \left(\frac{x-1}{x+1}\right) + C$}$.

The integral we need to evaluate is $\int\limits x \sin x \, dx$.

We can solve this integral using the method of Integration by Parts.

The formula for Integration by Parts is given by:

$\int\limits u \, dv = uv - \int\limits v \, du$

In the given integral $\int\limits x \sin x \, dx$, we need to choose appropriate functions for $u$ and $dv$. A common guideline for choosing $u$ is the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential).

Following the LIATE rule, we let:

$u = x$ (Algebraic function)

$dv = \sin x \, dx$ (Trigonometric function)

Now, we find $du$ by differentiating $u$ with respect to $x$, and find $v$ by integrating $dv$ with respect to $x$.

$du = \frac{d}{dx}(x) \, dx = 1 \cdot dx = dx$

$v = \int\limits \sin x \, dx = -\cos x$

Substitute these values into the Integration by Parts formula $\int\limits u \, dv = uv - \int\limits v \, du$:

$\int\limits x \sin x \, dx = (x)(-\cos x) - \int\limits (-\cos x) \, dx$

$\int\limits x \sin x \, dx = -x \cos x - \int\limits (-\cos x) \, dx$

$\int\limits x \sin x \, dx = -x \cos x + \int\limits \cos x \, dx$

Now, we evaluate the remaining integral $\int\limits \cos x \, dx$:

$\int\limits \cos x \, dx = \sin x$

Substitute the result of the second integral back into the equation obtained from the Integration by Parts formula:

$\int\limits x \sin x \, dx = -x \cos x + \sin x + C$

Here, $C$ is the constant of integration, which is added to account for the indefinite nature of the integral.

Final Result:

The integral of $x \sin x$ with respect to $x$ is $-x \cos x + \sin x + C$.

$\int\limits x \sin x \, dx = -x \cos x + \sin x + C$

We are asked to find the integral of the function $x \sin 3x$ with respect to $x$. The integral is $\int\limits x \sin 3x \, dx$.

We will use the method of Integration by Parts to solve this integral.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

We need to choose suitable functions for $u$ and $dv$. Using the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) as a guide, we choose:

$u = x$ (Algebraic function)

$dv = \sin 3x \, dx$ (Trigonometric function)

Now, we differentiate $u$ to find $du$ and integrate $dv$ to find $v$.

Differentiating $u$ with respect to $x$:

$du = \frac{d}{dx}(x) \, dx = 1 \cdot dx = dx$

Integrating $dv$ with respect to $x$:

$v = \int\limits \sin 3x \, dx$

To integrate $\sin 3x$, we use the substitution method or recall the standard integral form $\int \sin(ax) \, dx = -\frac{1}{a}\cos(ax)$. Here $a=3$.

$v = -\frac{1}{3} \cos 3x$

Substitute these values of $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits x \sin 3x \, dx = (x)\left(-\frac{1}{3}\cos 3x\right) - \int\limits \left(-\frac{1}{3}\cos 3x\right) \, dx$

$\int\limits x \sin 3x \, dx = -\frac{x}{3}\cos 3x - \int\limits -\frac{1}{3}\cos 3x \, dx$

$\int\limits x \sin 3x \, dx = -\frac{x}{3}\cos 3x + \frac{1}{3}\int\limits \cos 3x \, dx$

Now, we evaluate the remaining integral $\int\limits \cos 3x \, dx$. We use the standard integral form $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax)$. Here $a=3$.

$\int\limits \cos 3x \, dx = \frac{1}{3}\sin 3x$

Substitute this result back into the equation from the Integration by Parts formula:

$\int\limits x \sin 3x \, dx = -\frac{x}{3}\cos 3x + \frac{1}{3}\left(\frac{1}{3}\sin 3x\right) + C$

$\int\limits x \sin 3x \, dx = -\frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + C$

Here, $C$ is the constant of integration.

Final Answer:

The integral of $x \sin 3x$ is $-\frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + C$.

$\int\limits x \sin 3x \, dx = -\frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + C$

We need to evaluate the integral $\int\limits x^2 e^x \, dx$.

This integral can be solved using Integration by Parts.

The Integration by Parts formula is:

$\int\limits u \, dv = uv - \int\limits v \, du$

We choose $u$ and $dv$ according to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential).

Let $u = x^2$ (Algebraic) and $dv = e^x \, dx$ (Exponential).

Differentiate $u$ to find $du$:

$du = \frac{d}{dx}(x^2) \, dx = 2x \, dx$

Integrate $dv$ to find $v$:

$v = \int\limits e^x \, dx = e^x$

Apply the Integration by Parts formula:

$\int\limits x^2 e^x \, dx = (x^2)(e^x) - \int\limits (e^x)(2x) \, dx$

$\int\limits x^2 e^x \, dx = x^2 e^x - 2 \int\limits x e^x \, dx$

Now, we need to evaluate the new integral $\int\limits x e^x \, dx$. This also requires Integration by Parts.

For the integral $\int\limits x e^x \, dx$, let $u_1 = x$ and $dv_1 = e^x \, dx$ (again following LIATE).

Differentiate $u_1$ to find $du_1$:

$du_1 = \frac{d}{dx}(x) \, dx = 1 \, dx = dx$

Integrate $dv_1$ to find $v_1$:

$v_1 = \int\limits e^x \, dx = e^x$

Apply the Integration by Parts formula to $\int\limits x e^x \, dx$:

$\int\limits x e^x \, dx = (x)(e^x) - \int\limits (e^x)(dx)$

$\int\limits x e^x \, dx = x e^x - \int\limits e^x \, dx$

$\int\limits x e^x \, dx = x e^x - e^x + C_1$

Substitute the result of $\int\limits x e^x \, dx$ back into the main equation:

$\int\limits x^2 e^x \, dx = x^2 e^x - 2 (x e^x - e^x) + C$

$\int\limits x^2 e^x \, dx = x^2 e^x - 2x e^x + 2e^x + C$

where $C$ is the constant of integration.

We can factor out $e^x$ from the terms:

$\int\limits x^2 e^x \, dx = e^x (x^2 - 2x + 2) + C$

Final Answer:

The integral of $x^2 e^x$ is $e^x (x^2 - 2x + 2) + C$.

$\int\limits x^2 e^x \, dx = e^x (x^2 - 2x + 2) + C$

We are asked to evaluate the integral $\int\limits x \log x \, dx$.

We will use the method of Integration by Parts.

The formula for Integration by Parts is given by:

$\int\limits u \, dv = uv - \int\limits v \, du$

We need to choose the functions for $u$ and $dv$. According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), Logarithmic functions are chosen as $u$ before Algebraic functions.

Therefore, we set:

$u = \log x$ (Logarithmic function)

$dv = x \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$, and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$:

$du = \frac{d}{dx}(\log x) \, dx = \frac{1}{x} \, dx$

Integrate $dv$:

$v = \int\limits x \, dx = \frac{x^2}{2}$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula $\int\limits u \, dv = uv - \int\limits v \, du$:

$\int\limits x \log x \, dx = (\log x)\left(\frac{x^2}{2}\right) - \int\limits \left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) \, dx$

$\int\limits x \log x \, dx = \frac{x^2}{2} \log x - \int\limits \frac{x}{2} \, dx$

Now, we evaluate the remaining integral $\int\limits \frac{x}{2} \, dx$:

$\int\limits \frac{x}{2} \, dx = \frac{1}{2} \int\limits x \, dx = \frac{1}{2} \left(\frac{x^2}{2}\right) + C'$

$\int\limits \frac{x}{2} \, dx = \frac{x^2}{4} + C'$

Substitute the result back into the main equation:

$\int\limits x \log x \, dx = \frac{x^2}{2} \log x - \left(\frac{x^2}{4}\right) + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $x \log x$ is $\frac{x^2}{2} \log x - \frac{x^2}{4} + C$.

$\int\limits x \log x \, dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + C$

We need to evaluate the integral $\int\limits x \log (2x) \, dx$.

We will use the method of Integration by Parts.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

Following the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we choose $u$ and $dv$ as follows:

$u = \log(2x)$ (Logarithmic function)

$dv = x \, dx$ (Algebraic function)

Now, we differentiate $u$ to find $du$ and integrate $dv$ to find $v$.

Differentiating $u$ with respect to $x$:

$du = \frac{d}{dx}(\log(2x)) \, dx$

Using the chain rule, $\frac{d}{dx}(\log(ax)) = \frac{1}{ax} \cdot a = \frac{1}{x}$.

$du = \frac{1}{2x} \cdot 2 \, dx = \frac{1}{x} \, dx$

Integrating $dv$ with respect to $x$:

$v = \int\limits x \, dx = \frac{x^2}{2}$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula $\int\limits u \, dv = uv - \int\limits v \, du$:

$\int\limits x \log(2x) \, dx = (\log(2x))\left(\frac{x^2}{2}\right) - \int\limits \left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) \, dx$

$\int\limits x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \int\limits \frac{x}{2} \, dx$

Now, we evaluate the remaining integral $\int\limits \frac{x}{2} \, dx$:

$\int\limits \frac{x}{2} \, dx = \frac{1}{2} \int\limits x \, dx = \frac{1}{2} \left(\frac{x^2}{2}\right) = \frac{x^2}{4}$

Substitute this result back into the main equation:

$\int\limits x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \frac{x^2}{4} + C$

where $C$ is the constant of integration.

We can also write $\log(2x) = \log 2 + \log x$ and integrate term by term, but the above method is direct.

Final Answer:

The integral of $x \log 2x$ is $\frac{x^2}{2} \log(2x) - \frac{x^2}{4} + C$.

$\int\limits x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \frac{x^2}{4} + C$

We need to evaluate the integral $\int\limits x^2 \log x \, dx$.

This integral can be solved using the method of Integration by Parts.

The Integration by Parts formula is:

$\int\limits u \, dv = uv - \int\limits v \, du$

We choose $u$ and $dv$ based on the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). The Logarithmic function comes before the Algebraic function.

Let $u = \log x$ (Logarithmic function)

Let $dv = x^2 \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiating $u$:

$du = \frac{d}{dx}(\log x) \, dx = \frac{1}{x} \, dx$

Integrating $dv$:

$v = \int\limits x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits x^2 \log x \, dx = (\log x)\left(\frac{x^3}{3}\right) - \int\limits \left(\frac{x^3}{3}\right)\left(\frac{1}{x}\right) \, dx$

$\int\limits x^2 \log x \, dx = \frac{x^3}{3} \log x - \int\limits \frac{x^3}{3x} \, dx$

$\int\limits x^2 \log x \, dx = \frac{x^3}{3} \log x - \int\limits \frac{x^2}{3} \, dx$

Now, we evaluate the remaining integral $\int\limits \frac{x^2}{3} \, dx$.

$\int\limits \frac{x^2}{3} \, dx = \frac{1}{3} \int\limits x^2 \, dx = \frac{1}{3} \left(\frac{x^3}{3}\right) = \frac{x^3}{9}$

Substitute this result back into the main equation:

$\int\limits x^2 \log x \, dx = \frac{x^3}{3} \log x - \frac{x^3}{9} + C$

where $C$ is the constant of integration.

We can factor out $\frac{x^3}{9}$ if desired:

$\int\limits x^2 \log x \, dx = \frac{x^3}{9} (3 \log x - 1) + C$

Final Answer:

The integral of $x^2 \log x$ is $\frac{x^3}{3} \log x - \frac{x^3}{9} + C$.

$\int\limits x^2 \log x \, dx = \frac{x^3}{3} \log x - \frac{x^3}{9} + C$

We need to evaluate the integral $\int\limits x \sin^{-1} x \, dx$.

We will solve this integral using the method of Integration by Parts.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

We use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to select $u$ and $dv$. The inverse trigonometric function comes before the algebraic function.

Let $u = \sin^{-1} x$ (Inverse trigonometric function)

Let $dv = x \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$:

$du = \frac{d}{dx}(\sin^{-1} x) \, dx = \frac{1}{\sqrt{1-x^2}} \, dx$

Integrate $dv$:

$v = \int\limits x \, dx = \frac{x^2}{2}$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits x \sin^{-1} x \, dx = (\sin^{-1} x)\left(\frac{x^2}{2}\right) - \int\limits \left(\frac{x^2}{2}\right)\left(\frac{1}{\sqrt{1-x^2}}\right) \, dx$

$\int\limits x \sin^{-1} x \, dx = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \int\limits \frac{x^2}{\sqrt{1-x^2}} \, dx$

Now, we need to evaluate the remaining integral $I' = \int\limits \frac{x^2}{\sqrt{1-x^2}} \, dx$. We can use a trigonometric substitution.

Let $x = \sin \theta$. Then $dx = \cos \theta \, d\theta$.

Also, $\sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = |\cos \theta|$. For the standard range of $\sin^{-1}x$ ($-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$), $\cos \theta \geq 0$, so $\sqrt{1-x^2} = \cos \theta$.

Substitute these into the integral $I'$:

$I' = \int\limits \frac{(\sin \theta)^2}{\cos \theta} (\cos \theta \, d\theta) = \int\limits \sin^2 \theta \, d\theta$

Use the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:

$I' = \int\limits \frac{1 - \cos 2\theta}{2} \, d\theta = \frac{1}{2} \int\limits (1 - \cos 2\theta) \, d\theta$

$I' = \frac{1}{2} \left(\int\limits 1 \, d\theta - \int\limits \cos 2\theta \, d\theta\right)$

$I' = \frac{1}{2} \left(\theta - \frac{\sin 2\theta}{2}\right) + C'$

Use the identity $\sin 2\theta = 2 \sin \theta \cos \theta$:

$I' = \frac{1}{2} \left(\theta - \frac{2 \sin \theta \cos \theta}{2}\right) + C' = \frac{1}{2} (\theta - \sin \theta \cos \theta) + C'$

Now, substitute back in terms of $x$. Recall $x = \sin \theta$, so $\theta = \sin^{-1} x$. Also, $\cos \theta = \sqrt{1-x^2}$.

$I' = \frac{1}{2} (\sin^{-1} x - x\sqrt{1-x^2}) + C'$

Substitute the value of $I'$ back into the main integral equation:

$\int\limits x \sin^{-1} x \, dx = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \left[\frac{1}{2} (\sin^{-1} x - x\sqrt{1-x^2})\right] + C$

$\int\limits x \sin^{-1} x \, dx = \frac{x^2}{2} \sin^{-1} x - \frac{1}{4} \sin^{-1} x + \frac{x\sqrt{1-x^2}}{4} + C$

Combine the terms with $\sin^{-1} x$:

$\int\limits x \sin^{-1} x \, dx = \sin^{-1} x \left(\frac{x^2}{2} - \frac{1}{4}\right) + \frac{x\sqrt{1-x^2}}{4} + C$

$\int\limits x \sin^{-1} x \, dx = \sin^{-1} x \left(\frac{2x^2 - 1}{4}\right) + \frac{x\sqrt{1-x^2}}{4} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $x \sin^{-1} x$ is $\frac{2x^2 - 1}{4} \sin^{-1} x + \frac{x\sqrt{1-x^2}}{4} + C$.

$\int\limits x \sin^{-1} x \, dx = \frac{2x^2 - 1}{4} \sin^{-1} x + \frac{x\sqrt{1-x^2}}{4} + C$

We need to evaluate the integral $\int\limits x \tan^{-1} x \, dx$.

We will solve this integral using the method of Integration by Parts.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

We use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to select $u$ and $dv$. The inverse trigonometric function ($\tan^{-1} x$) comes before the algebraic function ($x$).

Let $u = \tan^{-1} x$ (Inverse trigonometric function)

Let $dv = x \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$:

$du = \frac{d}{dx}(\tan^{-1} x) \, dx = \frac{1}{1+x^2} \, dx$

Integrate $dv$:

$v = \int\limits x \, dx = \frac{x^2}{2}$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits x \tan^{-1} x \, dx = (\tan^{-1} x)\left(\frac{x^2}{2}\right) - \int\limits \left(\frac{x^2}{2}\right)\left(\frac{1}{1+x^2}\right) \, dx$

$\int\limits x \tan^{-1} x \, dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int\limits \frac{x^2}{1+x^2} \, dx$

Now, we need to evaluate the remaining integral $I' = \int\limits \frac{x^2}{1+x^2} \, dx$.

We can rewrite the integrand by adding and subtracting 1 in the numerator:

$\frac{x^2}{1+x^2} = \frac{x^2 + 1 - 1}{1+x^2} = \frac{1+x^2}{1+x^2} - \frac{1}{1+x^2} = 1 - \frac{1}{1+x^2}$

Now, integrate $I'$:

$I' = \int\limits \left(1 - \frac{1}{1+x^2}\right) \, dx = \int\limits 1 \, dx - \int\limits \frac{1}{1+x^2} \, dx$

The integral of $1$ is $x$, and the integral of $\frac{1}{1+x^2}$ is $\tan^{-1} x$.

$I' = x - \tan^{-1} x + C'$

Substitute the value of $I'$ back into the main integral equation:

$\int\limits x \tan^{-1} x \, dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} (x - \tan^{-1} x) + C$

$\int\limits x \tan^{-1} x \, dx = \frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x + C$

Combine the terms involving $\tan^{-1} x$:

$\int\limits x \tan^{-1} x \, dx = \left(\frac{x^2}{2} + \frac{1}{2}\right) \tan^{-1} x - \frac{x}{2} + C$

$\int\limits x \tan^{-1} x \, dx = \frac{x^2+1}{2} \tan^{-1} x - \frac{x}{2} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $x \tan^{-1} x$ is $\frac{x^2+1}{2} \tan^{-1} x - \frac{x}{2} + C$.

$\int\limits x \tan^{-1} x \, dx = \frac{x^2+1}{2} \tan^{-1} x - \frac{x}{2} + C$

We need to evaluate the integral $\int\limits x \cos^{-1} x \, dx$.

We will solve this integral using the method of Integration by Parts.

The formula for Integration by Parts is given by:

$\int\limits u \, dv = uv - \int\limits v \, du$

We use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to select $u$ and $dv$. The inverse trigonometric function ($\cos^{-1} x$) comes before the algebraic function ($x$).

Let $u = \cos^{-1} x$ (Inverse trigonometric function)

Let $dv = x \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$:

$du = \frac{d}{dx}(\cos^{-1} x) \, dx = -\frac{1}{\sqrt{1-x^2}} \, dx$

Integrate $dv$:

$v = \int\limits x \, dx = \frac{x^2}{2}$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits x \cos^{-1} x \, dx = (\cos^{-1} x)\left(\frac{x^2}{2}\right) - \int\limits \left(\frac{x^2}{2}\right)\left(-\frac{1}{\sqrt{1-x^2}}\right) \, dx$

$\int\limits x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x - \int\limits -\frac{x^2}{2\sqrt{1-x^2}} \, dx$

$\int\limits x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \int\limits \frac{x^2}{\sqrt{1-x^2}} \, dx$

Now, we need to evaluate the remaining integral $I' = \int\limits \frac{x^2}{\sqrt{1-x^2}} \, dx$. We use a trigonometric substitution.

Let $x = \sin \theta$. Then $dx = \cos \theta \, d\theta$.

Also, $\sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = |\cos \theta|$. For $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, $\cos \theta \geq 0$, so $\sqrt{1-x^2} = \cos \theta$. The substitution $x = \sin \theta$ is valid for the domain of $\cos^{-1} x$ ($[-1, 1]$ which corresponds to $0 \leq \theta \leq \pi$), but $\sin^{-1}x$ is often used for this integral and its standard domain is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. We proceed with $x = \sin \theta$ and $\sqrt{1-x^2} = \cos \theta$ (assuming appropriate range).

Substitute these into the integral $I'$:

$I' = \int\limits \frac{(\sin \theta)^2}{\cos \theta} (\cos \theta \, d\theta) = \int\limits \sin^2 \theta \, d\theta$

Use the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:

$I' = \int\limits \frac{1 - \cos 2\theta}{2} \, d\theta = \frac{1}{2} \int\limits (1 - \cos 2\theta) \, d\theta$

$I' = \frac{1}{2} \left(\int\limits 1 \, d\theta - \int\limits \cos 2\theta \, d\theta\right) = \frac{1}{2} \left(\theta - \frac{\sin 2\theta}{2}\right) + C'$

Use the identity $\sin 2\theta = 2 \sin \theta \cos \theta$:

$I' = \frac{1}{2} (\theta - \sin \theta \cos \theta) + C'$

Now, substitute back in terms of $x$. Recall $x = \sin \theta$, so $\theta = \sin^{-1} x$. Also, $\cos \theta = \sqrt{1-x^2}$.

$I' = \frac{1}{2} (\sin^{-1} x - x\sqrt{1-x^2})$ (excluding $C'$ here)

Substitute the value of $I'$ back into the main integral equation:

$\int\limits x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \left[\frac{1}{2} (\sin^{-1} x - x\sqrt{1-x^2})\right] + C$

$\int\limits x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x + \frac{1}{4} \sin^{-1} x - \frac{x\sqrt{1-x^2}}{4} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $x \cos^{-1} x$ is $\frac{x^2}{2} \cos^{-1} x + \frac{1}{4} \sin^{-1} x - \frac{x\sqrt{1-x^2}}{4} + C$.

$\int\limits x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x + \frac{1}{4} \sin^{-1} x - \frac{x\sqrt{1-x^2}}{4} + C$

We need to evaluate the integral $\int\limits (\sin^{-1} x)^2 \, dx$.

We will solve this integral using the method of Integration by Parts. Since the integrand is a single function, we can consider $dv = dx$.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

Let $u = (\sin^{-1} x)^2$

Let $dv = dx$

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$ using the chain rule:

$du = \frac{d}{dx}((\sin^{-1} x)^2) \, dx = 2(\sin^{-1} x) \cdot \frac{d}{dx}(\sin^{-1} x) \, dx$

$du = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} \, dx = \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} \, dx$

Integrate $dv$:

$v = \int\limits dx = x$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits (\sin^{-1} x)^2 \, dx = (\sin^{-1} x)^2 (x) - \int\limits x \cdot \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} \, dx$

$\int\limits (\sin^{-1} x)^2 \, dx = x (\sin^{-1} x)^2 - 2 \int\limits \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \, dx$

Now, we need to evaluate the new integral $I' = \int\limits \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \, dx$. We can use a substitution.

Let $w = \sin^{-1} x$. Then $dw = \frac{1}{\sqrt{1-x^2}} \, dx$. Also, $x = \sin w$.

Substitute these into the integral $I'$:

$I' = \int\limits x \cdot \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \, dx$

$I' = \int\limits (\sin w) \cdot (w) \, dw = \int\limits w \sin w \, dw$

The integral $\int\limits w \sin w \, dw$ is of the form $\int\limits u_1 \, dv_1$ which requires Integration by Parts again. We use the LIATE rule for this new integral (considering 'w' as algebraic and 'sin w' as trigonometric).

Let $u_1 = w$

Let $dv_1 = \sin w \, dw$

Differentiate $u_1$ to find $du_1$ and integrate $dv_1$ to find $v_1$.

$du_1 = \frac{d}{dw}(w) \, dw = 1 \cdot dw = dw$

$v_1 = \int\limits \sin w \, dw = -\cos w$

Apply the Integration by Parts formula to $\int\limits w \sin w \, dw$:

$I' = (w)(-\cos w) - \int\limits (-\cos w) \, dw$

$I' = -w \cos w + \int\limits \cos w \, dw$

$I' = -w \cos w + \sin w$

Now, substitute back from $w$ to $x$. Recall $w = \sin^{-1} x$ and $x = \sin w$. To find $\cos w$ in terms of $x$, we use the identity $\cos w = \sqrt{1-\sin^2 w} = \sqrt{1-x^2}$ (considering the principal value range of $\sin^{-1}x$ where $\cos w \geq 0$).

$I' = -(\sin^{-1} x)\sqrt{1-x^2} + x$

Substitute the result of $I'$ back into the main integral equation:

$\int\limits (\sin^{-1} x)^2 \, dx = x (\sin^{-1} x)^2 - 2 \left[ -(\sin^{-1} x)\sqrt{1-x^2} + x \right] + C$

$\int\limits (\sin^{-1} x)^2 \, dx = x (\sin^{-1} x)^2 + 2 (\sin^{-1} x)\sqrt{1-x^2} - 2x + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $(\sin^{-1} x)^2$ is $x (\sin^{-1} x)^2 + 2 \sqrt{1-x^2} \sin^{-1} x - 2x + C$.

$\int\limits (\sin^{-1} x)^2 \, dx = x (\sin^{-1} x)^2 + 2 \sqrt{1-x^2} \sin^{-1} x - 2x + C$

We need to evaluate the integral $\int\limits \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \, dx$.

We observe that the derivative of $\cos^{-1} x$ is related to the term $\frac{1}{\sqrt{1-x^2}}$. This suggests using a substitution.

Let's use the substitution method.

Let $u = \cos^{-1} x$.

Differentiating both sides with respect to $x$, we get:

$du = \frac{d}{dx}(\cos^{-1} x) \, dx = -\frac{1}{\sqrt{1-x^2}} \, dx$

So, $\frac{1}{\sqrt{1-x^2}} \, dx = -du$.

Also, from $u = \cos^{-1} x$, we have $x = \cos u$.

Now, substitute $u$, $x$, and $dx$ terms into the integral:

$\int\limits \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \, dx = \int\limits (\cos u) \cdot (u) \cdot (-du)$

$\int\limits \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \, dx = -\int\limits u \cos u \, du$

The resulting integral $-\int\limits u \cos u \, du$ requires Integration by Parts.

The Integration by Parts formula is:

$\int\limits U \, dV = UV - \int\limits V \, dU$

For the integral $\int\limits u \cos u \, du$, we choose $U$ and $dV$ using the LIATE rule (considering $u$ as algebraic and $\cos u$ as trigonometric):

Let $U = u$

Let $dV = \cos u \, du$

Now, we find $dU$ by differentiating $U$ and $V$ by integrating $dV$:

$dU = \frac{d}{du}(u) \, du = 1 \cdot du = du$

$V = \int\limits \cos u \, du = \sin u$

Apply the Integration by Parts formula to $\int\limits u \cos u \, du$:

$\int\limits u \cos u \, du = (u)(\sin u) - \int\limits (\sin u) \, du$

$\int\limits u \cos u \, du = u \sin u - \int\limits \sin u \, du$

Evaluate the remaining integral $\int\limits \sin u \, du = -\cos u$.

So, $\int\limits u \cos u \, du = u \sin u - (-\cos u) + C'$

$\int\limits u \cos u \, du = u \sin u + \cos u + C'$

Now, substitute this result back into the expression for the original integral:

$\int\limits \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \, dx = -\int\limits u \cos u \, du = -(u \sin u + \cos u) + C$

$\int\limits \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \, dx = -u \sin u - \cos u + C$

Finally, substitute back from $u$ to $x$. Recall $u = \cos^{-1} x$, $x = \cos u$. For $\sin u$, we use the identity $\sin u = \sqrt{1 - \cos^2 u} = \sqrt{1 - x^2}$. Since the principal value range of $\cos^{-1} x$ is $[0, \pi]$, $\sin u = \sin(\cos^{-1} x)$ is non-negative, so the positive square root is appropriate.

Substitute $u = \cos^{-1} x$, $\sin u = \sqrt{1-x^2}$, and $\cos u = x$ into the expression:

$-(\cos^{-1} x)(\sqrt{1-x^2}) - x + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $\frac{x \cos^{-1} x}{\sqrt{1 - x^2}}$ is $-\sqrt{1-x^2} \cos^{-1} x - x + C$.

$\int\limits \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \, dx = -\sqrt{1-x^2} \cos^{-1} x - x + C$

We need to evaluate the integral $\int\limits x \sec^2 x \, dx$.

We will solve this integral using the method of Integration by Parts.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

We use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to select $u$ and $dv$. The algebraic function ($x$) comes before the trigonometric function ($\sec^2 x$).

Let $u = x$ (Algebraic function)

Let $dv = \sec^2 x \, dx$ (Trigonometric function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$:

$du = \frac{d}{dx}(x) \, dx = 1 \, dx = dx$

Integrate $dv$:

$v = \int\limits \sec^2 x \, dx = \tan x$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits x \sec^2 x \, dx = (x)(\tan x) - \int\limits (\tan x) \, dx$

$\int\limits x \sec^2 x \, dx = x \tan x - \int\limits \tan x \, dx$

Now, we evaluate the remaining integral $\int\limits \tan x \, dx$.

The standard integral of $\tan x$ is $\log|\sec x|$.

$\int\limits \tan x \, dx = \log|\sec x|$

Substitute this result back into the main equation:

$\int\limits x \sec^2 x \, dx = x \tan x - \log|\sec x| + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $x \sec^2 x$ is $x \tan x - \log|\sec x| + C$.

$\int\limits x \sec^2 x \, dx = x \tan x - \log|\sec x| + C$

We need to evaluate the integral $\int\limits \tan^{-1} x \, dx$.

Since the integrand is a single inverse trigonometric function, we can use the method of Integration by Parts by considering the integrand as $\tan^{-1} x \cdot 1$.

The formula for Integration by Parts is given by:

$\int\limits u \, dv = uv - \int\limits v \, du$

We choose $u$ and $dv$. According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), the inverse trigonometric function comes first.

Let $u = \tan^{-1} x$ (Inverse trigonometric function)

Let $dv = 1 \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$:

$du = \frac{d}{dx}(\tan^{-1} x) \, dx = \frac{1}{1+x^2} \, dx$

Integrate $dv$:

$v = \int\limits 1 \, dx = x$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits \tan^{-1} x \, dx = (\tan^{-1} x)(x) - \int\limits (x)\left(\frac{1}{1+x^2}\right) \, dx$

$\int\limits \tan^{-1} x \, dx = x \tan^{-1} x - \int\limits \frac{x}{1+x^2} \, dx$

Now, we need to evaluate the remaining integral $I' = \int\limits \frac{x}{1+x^2} \, dx$. We can use a substitution method.

Let $w = 1+x^2$.

Differentiate $w$ with respect to $x$: $\frac{dw}{dx} = 2x$.

So, $dw = 2x \, dx$, which means $x \, dx = \frac{1}{2} dw$.

Substitute into the integral $I'$:

$I' = \int\limits \frac{1}{1+x^2} (x \, dx) = \int\limits \frac{1}{w} \left(\frac{1}{2} dw\right)$

$I' = \frac{1}{2} \int\limits \frac{1}{w} \, dw$

The integral of $\frac{1}{w}$ is $\log|w|$.

$I' = \frac{1}{2} \log|w| + C'$

Substitute back $w = 1+x^2$. Since $1+x^2 > 0$ for all real $x$, we can remove the absolute value.

$I' = \frac{1}{2} \log(1+x^2) + C'$

Substitute the value of $I'$ back into the main integral equation:

$\int\limits \tan^{-1} x \, dx = x \tan^{-1} x - \left(\frac{1}{2} \log(1+x^2)\right) + C$

$\int\limits \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \log(1+x^2) + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $\tan^{-1} x$ is $x \tan^{-1} x - \frac{1}{2} \log(1+x^2) + C$.

$\int\limits \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \log(1+x^2) + C$

We need to evaluate the integral $\int\limits x (\log x)^2 \, dx$.

We will solve this integral using the method of Integration by Parts.

The formula for Integration by Parts is given by:

$\int\limits u \, dv = uv - \int\limits v \, du$

We choose $u$ and $dv$ based on the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). The logarithmic function comes before the algebraic function.

Let $u = (\log x)^2$ (Logarithmic function)

Let $dv = x \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$ using the chain rule:

$du = \frac{d}{dx}((\log x)^2) \, dx = 2(\log x) \cdot \frac{d}{dx}(\log x) \, dx$

$du = 2(\log x) \cdot \frac{1}{x} \, dx = \frac{2 \log x}{x} \, dx$

Integrate $dv$:

$v = \int\limits x \, dx = \frac{x^2}{2}$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits x (\log x)^2 \, dx = ((\log x)^2)\left(\frac{x^2}{2}\right) - \int\limits \left(\frac{x^2}{2}\right)\left(\frac{2 \log x}{x}\right) \, dx$

$\int\limits x (\log x)^2 \, dx = \frac{x^2}{2} (\log x)^2 - \int\limits \frac{2x^2 \log x}{2x} \, dx$

$\int\limits x (\log x)^2 \, dx = \frac{x^2}{2} (\log x)^2 - \int\limits x \log x \, dx$

Now, we need to evaluate the new integral $I' = \int\limits x \log x \, dx$. This integral also requires Integration by Parts.

For the integral $\int\limits x \log x \, dx$, we again use the LIATE rule. Let $u_1 = \log x$ and $dv_1 = x \, dx$.

Differentiate $u_1$ to find $du_1$ and integrate $dv_1$ to find $v_1$.

$du_1 = \frac{d}{dx}(\log x) \, dx = \frac{1}{x} \, dx$

$v_1 = \int\limits x \, dx = \frac{x^2}{2}$

Apply the Integration by Parts formula to $\int\limits x \log x \, dx = u_1 v_1 - \int\limits v_1 \, du_1$:

$\int\limits x \log x \, dx = (\log x)\left(\frac{x^2}{2}\right) - \int\limits \left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) \, dx$

$\int\limits x \log x \, dx = \frac{x^2}{2} \log x - \int\limits \frac{x}{2} \, dx$

Now evaluate $\int\limits \frac{x}{2} \, dx = \frac{1}{2} \int\limits x \, dx = \frac{1}{2} \left(\frac{x^2}{2}\right) = \frac{x^2}{4}$.

So, $\int\limits x \log x \, dx = \frac{x^2}{2} \log x - \frac{x^2}{4}$ (we'll add the constant $C$ at the end).

Substitute the result of $I'$ back into the main integral equation:

$\int\limits x (\log x)^2 \, dx = \frac{x^2}{2} (\log x)^2 - \left( \frac{x^2}{2} \log x - \frac{x^2}{4} \right) + C$

$\int\limits x (\log x)^2 \, dx = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $x (\log x)^2$ is $\frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C$.

$\int\limits x (\log x)^2 \, dx = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C$

We need to evaluate the integral $\int\limits (x^2 + 1) \log x \, dx$.

We will use the method of Integration by Parts.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we choose the logarithmic function as $u$ and the algebraic function as $dv$.

Let $u = \log x$ (Logarithmic function)

Let $dv = (x^2 + 1) \, dx$ (Algebraic function)

Now, we find $du$ by differentiating $u$ with respect to $x$ and $v$ by integrating $dv$ with respect to $x$.

Differentiate $u$:

$du = \frac{d}{dx}(\log x) \, dx = \frac{1}{x} \, dx$

Integrate $dv$:

$v = \int\limits (x^2 + 1) \, dx = \int\limits x^2 \, dx + \int\limits 1 \, dx = \frac{x^3}{3} + x$

Substitute $u$, $v$, and $du$ into the Integration by Parts formula:

$\int\limits (x^2 + 1) \log x \, dx = (\log x)\left(\frac{x^3}{3} + x\right) - \int\limits \left(\frac{x^3}{3} + x\right)\left(\frac{1}{x}\right) \, dx$

$\int\limits (x^2 + 1) \log x \, dx = \left(\frac{x^3}{3} + x\right) \log x - \int\limits \left(\frac{x^3}{3x} + \frac{x}{x}\right) \, dx$

$\int\limits (x^2 + 1) \log x \, dx = \left(\frac{x^3}{3} + x\right) \log x - \int\limits \left(\frac{x^2}{3} + 1\right) \, dx$

Now, we evaluate the remaining integral $\int\limits \left(\frac{x^2}{3} + 1\right) \, dx$.

$\int\limits \left(\frac{x^2}{3} + 1\right) \, dx = \int\limits \frac{x^2}{3} \, dx + \int\limits 1 \, dx$

$\int\limits \frac{x^2}{3} \, dx = \frac{1}{3} \int\limits x^2 \, dx = \frac{1}{3} \left(\frac{x^3}{3}\right) = \frac{x^3}{9}$

$\int\limits 1 \, dx = x$

So, $\int\limits \left(\frac{x^2}{3} + 1\right) \, dx = \frac{x^3}{9} + x$ (we'll add the constant $C$ at the end).

Substitute this result back into the main equation:

$\int\limits (x^2 + 1) \log x \, dx = \left(\frac{x^3}{3} + x\right) \log x - \left(\frac{x^3}{9} + x\right) + C$

$\int\limits (x^2 + 1) \log x \, dx = \left(\frac{x^3}{3} + x\right) \log x - \frac{x^3}{9} - x + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $(x^2 + 1) \log x$ is $\left(\frac{x^3}{3} + x\right) \log x - \frac{x^3}{9} - x + C$.

$\int\limits (x^2 + 1) \log x \, dx = \left(\frac{x^3}{3} + x\right) \log x - \frac{x^3}{9} - x + C$

We need to evaluate the integral $\int\limits e^x (\sin x + \cos x) \, dx$.

This integral is of the standard form $\int\limits e^x [f(x) + f'(x)] \, dx$.

The integral of this form is given by:

$\int\limits e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$

In the given integral, $\int\limits e^x (\sin x + \cos x) \, dx$, we can identify $f(x)$ and $f'(x)$.

Let $f(x) = \sin x$.

Then the derivative of $f(x)$ is $f'(x) = \frac{d}{dx}(\sin x) = \cos x$.

The integrand is $e^x (\sin x + \cos x)$, which perfectly matches the form $e^x [f(x) + f'(x)]$ with $f(x) = \sin x$ and $f'(x) = \cos x$.

Using the standard formula, the integral is:

$\int\limits e^x (\sin x + \cos x) \, dx = e^x (\sin x) + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $e^x (\sin x + \cos x)$ is $e^x \sin x + C$.

$\int\limits e^x (\sin x + \cos x) \, dx = e^x \sin x + C$

Alternative Method (using Integration by Parts):

We can also evaluate this by splitting the integral and using Integration by Parts on one part.

Let $I = \int\limits e^x (\sin x + \cos x) \, dx = \int\limits e^x \sin x \, dx + \int\limits e^x \cos x \, dx$.

Let's evaluate $\int\limits e^x \sin x \, dx$ using Integration by Parts. Using LIATE, let $u = \sin x$ and $dv = e^x \, dx$.

$du = \cos x \, dx$

$v = e^x$

$\int\limits e^x \sin x \, dx = (\sin x)(e^x) - \int\limits (e^x)(\cos x) \, dx$

$\int\limits e^x \sin x \, dx = e^x \sin x - \int\limits e^x \cos x \, dx$

Substitute this back into the original integral $I$:

$I = \left(e^x \sin x - \int\limits e^x \cos x \, dx\right) + \int\limits e^x \cos x \, dx$

$I = e^x \sin x - \int\limits e^x \cos x \, dx + \int\limits e^x \cos x \, dx$

The two integral terms cancel out.

$I = e^x \sin x + C$

where $C$ is the constant of integration.

This confirms the result obtained using the standard form.

We need to evaluate the integral $\int\limits \frac{x e^x}{(1+x)^2} \, dx$.

We can rewrite the integrand to see if it fits the form $\int\limits e^x [f(x) + f'(x)] \, dx$.

Let's manipulate the numerator $x e^x$. We want to relate it to $(1+x)$.

$\frac{x}{(1+x)^2} = \frac{(1+x) - 1}{(1+x)^2} = \frac{1+x}{(1+x)^2} - \frac{1}{(1+x)^2} = \frac{1}{1+x} - \frac{1}{(1+x)^2}$

So, the integrand can be written as:

$\frac{x e^x}{(1+x)^2} = e^x \left(\frac{1}{1+x} - \frac{1}{(1+x)^2}\right)$

Now, the integral is $\int\limits e^x \left(\frac{1}{1+x} - \frac{1}{(1+x)^2}\right) \, dx$.

This integral is in the standard form $\int\limits e^x [f(x) + f'(x)] \, dx$.

Let $f(x) = \frac{1}{1+x} = (1+x)^{-1}$.

Let's find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}((1+x)^{-1})$.

Using the chain rule, $f'(x) = -1 \cdot (1+x)^{-1-1} \cdot \frac{d}{dx}(1+x) = -1 \cdot (1+x)^{-2} \cdot 1 = -\frac{1}{(1+x)^2}$.

So, $f(x) = \frac{1}{1+x}$ and $f'(x) = -\frac{1}{(1+x)^2}$.

The integrand is $e^x [f(x) + f'(x)]$.

Using the formula $\int\limits e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$, we have:

$\int\limits e^x \left(\frac{1}{1+x} - \frac{1}{(1+x)^2}\right) \, dx = e^x \cdot \frac{1}{1+x} + C$

$\int\limits \frac{x e^x}{(1+x)^2} \, dx = \frac{e^x}{1+x} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $\frac{x e^x}{(1+x)^2}$ is $\frac{e^x}{1+x} + C$.

$\int\limits \frac{x e^x}{(1+x)^2} \, dx = \frac{e^x}{1+x} + C$

We need to evaluate the integral $\int\limits e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) \, dx$.

We can try to express the term inside the parenthesis $\frac{1 + \sin x}{1 + \cos x}$ in the form $f(x) + f'(x)$, so that we can use the standard integral formula $\int\limits e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.

Let's simplify the expression $\frac{1 + \sin x}{1 + \cos x}$ using half-angle trigonometric identities.

We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$.

So, $\frac{1 + \sin x}{1 + \cos x} = \frac{1}{1 + \cos x} + \frac{\sin x}{1 + \cos x}$

Substitute the identities:

= $\frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}}$

= $\frac{1}{2} \cdot \frac{1}{\cos^2 \frac{x}{2}} + \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$

= $\frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2}$

So, the integrand is $e^x \left(\tan \frac{x}{2} + \frac{1}{2} \sec^2 \frac{x}{2}\right)$.

Now, we identify $f(x)$ and $f'(x)$ in the form $e^x [f(x) + f'(x)]$.

Let $f(x) = \tan \frac{x}{2}$.

Let's find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}\left(\tan \frac{x}{2}\right)$

Using the chain rule, $\frac{d}{dx}(\tan(ax)) = a \sec^2(ax)$. Here $a = \frac{1}{2}$.

$f'(x) = \frac{1}{2} \sec^2 \frac{x}{2}$.

We see that the expression $\left(\frac{1 + \sin x}{1 + \cos x}\right)$ is indeed equal to $\tan \frac{x}{2} + \frac{1}{2} \sec^2 \frac{x}{2}$, which is in the form $f(x) + f'(x)$ where $f(x) = \tan \frac{x}{2}$ and $f'(x) = \frac{1}{2} \sec^2 \frac{x}{2}$.

Using the standard integral formula $\int\limits e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$, we can write the integral directly:

$\int\limits e^x \left(\tan \frac{x}{2} + \frac{1}{2} \sec^2 \frac{x}{2}\right) \, dx = e^x \tan \frac{x}{2} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $e^x \left( \frac{1 + \sin x}{1 + \cos x} \right)$ is $e^x \tan \frac{x}{2} + C$.

$\int\limits e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) \, dx = e^x \tan \frac{x}{2} + C$

We need to evaluate the integral $\int\limits e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx$.

This integral is in the standard form $\int\limits e^x [f(x) + f'(x)] \, dx$.

The integral of this form is given by:

$\int\limits e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$

In the given integral, $\int\limits e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx$, we can identify $f(x)$ and $f'(x)$.

Let $f(x) = \frac{1}{x} = x^{-1}$.

Then the derivative of $f(x)$ is $f'(x) = \frac{d}{dx}(x^{-1})$.

$f'(x) = -1 \cdot x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$.

The expression inside the parenthesis is $\frac{1}{x} - \frac{1}{x^2}$, which is exactly $f(x) + f'(x)$ where $f(x) = \frac{1}{x}$ and $f'(x) = -\frac{1}{x^2}$.

Using the standard formula, the integral is:

$\int\limits e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx = e^x \cdot f(x) + C$

$\int\limits e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx = e^x \cdot \frac{1}{x} + C$

$\int\limits e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx = \frac{e^x}{x} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $e^x \left( \frac{1}{x} − \frac{1}{x^2} \right)$ is $\frac{e^x}{x} + C$.

$\int\limits e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx = \frac{e^x}{x} + C$

We need to evaluate the integral $\int\limits \frac{(x - 3) e^x}{(x - 1)^3} \, dx$.

We can rewrite the integrand to check if it fits the standard form $\int\limits e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.

Let's manipulate the fraction part $\frac{x - 3}{(x - 1)^3}$. We can express the numerator in terms of $(x-1)$.

$x - 3 = (x - 1) - 2$

Substitute this into the fraction:

$\frac{x - 3}{(x - 1)^3} = \frac{(x - 1) - 2}{(x - 1)^3}$

Now, split the fraction:

= $\frac{x - 1}{(x - 1)^3} - \frac{2}{(x - 1)^3}$

= $\frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3}$

So the integrand becomes $e^x \left(\frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3}\right)$.

We want to see if this is in the form $e^x [f(x) + f'(x)]$. Let's propose $f(x)$ and find its derivative.

Let $f(x) = \frac{1}{(x - 1)^2} = (x - 1)^{-2}$.

Now, let's find the derivative of $f(x)$ with respect to $x$, $f'(x)$:

$f'(x) = \frac{d}{dx}((x - 1)^{-2})$

Using the chain rule: $\frac{d}{du}(u^n) = nu^{n-1} \frac{du}{dx}$, with $u = x-1$ and $n = -2$.

$f'(x) = -2(x - 1)^{-2 - 1} \cdot \frac{d}{dx}(x - 1)$

$f'(x) = -2(x - 1)^{-3} \cdot 1$

$f'(x) = -\frac{2}{(x - 1)^3}$

We see that the integrand $e^x \left(\frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3}\right)$ is indeed of the form $e^x [f(x) + f'(x)]$ where $f(x) = \frac{1}{(x - 1)^2}$ and $f'(x) = -\frac{2}{(x - 1)^3}$.

Using the standard integral formula $\int\limits e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$, we get:

$\int\limits e^x \left(\frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3}\right) \, dx = e^x \cdot \frac{1}{(x - 1)^2} + C$

$\int\limits \frac{(x - 3) e^x}{(x - 1)^3} \, dx = \frac{e^x}{(x - 1)^2} + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $\frac{(x − 3) e^x}{(x − 1)^3}$ is $\frac{e^x}{(x − 1)^2} + C$.

$\int\limits \frac{(x - 3) e^x}{(x - 1)^3} \, dx = \frac{e^x}{(x - 1)^2} + C$

We need to evaluate the integral $\int\limits e^{2x} \sin x \, dx$.

This integral involves the product of an exponential function and a trigonometric function. It typically requires Integration by Parts applied twice.

The formula for Integration by Parts is:

$\int\limits u \, dv = uv - \int\limits v \, du$

Let the integral be $I = \int\limits e^{2x} \sin x \, dx$.

We can choose $u$ and $dv$ using the LIATE rule (though for exponential and trigonometric functions, the choice of which is $u$ doesn't affect whether it requires two steps, but it can influence the algebra). Let's choose the trigonometric function as $u$ first.

Let $u = \sin x$

Let $dv = e^{2x} \, dx$

Now, we find $du$ and $v$.

Differentiate $u$: $du = \frac{d}{dx}(\sin x) \, dx = \cos x \, dx$

Integrate $dv$: $v = \int\limits e^{2x} \, dx = \frac{e^{2x}}{2}$

Apply the Integration by Parts formula to $I$:

$I = (\sin x)\left(\frac{e^{2x}}{2}\right) - \int\limits \left(\frac{e^{2x}}{2}\right)(\cos x) \, dx$

$I = \frac{e^{2x}}{2} \sin x - \frac{1}{2} \int\limits e^{2x} \cos x \, dx$

Now, we need to evaluate the new integral $\int\limits e^{2x} \cos x \, dx$. Let's call this integral $I_1$. We apply Integration by Parts again to $I_1$.

For $I_1 = \int\limits e^{2x} \cos x \, dx$, we should maintain consistency in our choice of $u$ and $dv$. Since we chose the trigonometric function as $u$ in the first step, let's do the same here.

Let $u_1 = \cos x$

Let $dv_1 = e^{2x} \, dx$

Now, find $du_1$ and $v_1$.

Differentiate $u_1$: $du_1 = \frac{d}{dx}(\cos x) \, dx = -\sin x \, dx$

Integrate $dv_1$: $v_1 = \int\limits e^{2x} \, dx = \frac{e^{2x}}{2}$

Apply the Integration by Parts formula to $I_1 = \int\limits u_1 \, dv_1 = u_1 v_1 - \int\limits v_1 \, du_1$:

$I_1 = (\cos x)\left(\frac{e^{2x}}{2}\right) - \int\limits \left(\frac{e^{2x}}{2}\right)(-\sin x) \, dx$

$I_1 = \frac{e^{2x}}{2} \cos x + \frac{1}{2} \int\limits e^{2x} \sin x \, dx$

Notice that the integral on the right side is the original integral $I$. So we have:

$I_1 = \frac{e^{2x}}{2} \cos x + \frac{1}{2} I$

Substitute this expression for $I_1$ back into the equation for $I$:

$I = \frac{e^{2x}}{2} \sin x - \frac{1}{2} I_1$

$I = \frac{e^{2x}}{2} \sin x - \frac{1}{2} \left(\frac{e^{2x}}{2} \cos x + \frac{1}{2} I\right)$

$I = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x - \frac{1}{4} I$

Now, we have an equation where $I$ appears on both sides. Collect the $I$ terms on one side:

$I + \frac{1}{4} I = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x$

$\left(1 + \frac{1}{4}\right) I = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x$

$\frac{5}{4} I = \frac{e^{2x}}{4} (2 \sin x - \cos x)$

Solve for $I$ by multiplying both sides by $\frac{4}{5}$:

$I = \frac{4}{5} \cdot \frac{e^{2x}}{4} (2 \sin x - \cos x) + C$

$I = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $e^{2x} \sin x$ is $\frac{e^{2x}}{5} (2 \sin x - \cos x) + C$.

$\int\limits e^{2x} \sin x \, dx = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C$

We need to evaluate the integral $\int\limits \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \, dx$.

Let's simplify the term inside the inverse sine function using a trigonometric substitution.

We recognize the form $\frac{2x}{1+x^2}$ suggests the substitution $x = \tan \theta$.

Let $x = \tan \theta$. Then $dx = \sec^2 \theta \, d\theta$.

Substitute $x = \tan \theta$ into the argument of $\sin^{-1}$:

$\frac{2x}{1+x^2} = \frac{2 \tan \theta}{1 + \tan^2 \theta}$

Using the identity $1 + \tan^2 \theta = \sec^2 \theta$, we get:

= $\frac{2 \tan \theta}{\sec^2 \theta} = 2 \frac{\sin \theta}{\cos \theta} \cdot \cos^2 \theta = 2 \sin \theta \cos \theta$

Using the double angle identity $\sin(2\theta) = 2 \sin \theta \cos \theta$, we have:

= $\sin (2\theta)$

So, the integrand becomes $\sin^{-1}(\sin(2\theta))$.

For $x \in [-1, 1]$, which corresponds to $\theta \in [-\pi/4, \pi/4]$, $2\theta \in [-\pi/2, \pi/2]$. In this range, $\sin^{-1}(\sin y) = y$. Thus, $\sin^{-1}(\sin(2\theta)) = 2\theta$.

Since $x = \tan \theta$, we have $\theta = \tan^{-1} x$.

Therefore, for $x \in [-1, 1]$, the integrand simplifies to $2 \tan^{-1} x$.

The integral becomes $\int\limits 2 \tan^{-1} x \, dx = 2 \int\limits \tan^{-1} x \, dx$.

Now we need to evaluate the integral of $\tan^{-1} x$. We solved this using Integration by Parts in Question 13.

Let's apply Integration by Parts to $\int\limits \tan^{-1} x \, dx$:

Let $u = \tan^{-1} x$ and $dv = dx$.

$du = \frac{1}{1+x^2} \, dx$ and $v = x$.

$\int\limits \tan^{-1} x \, dx = x \tan^{-1} x - \int\limits x \cdot \frac{1}{1+x^2} \, dx$

$\int\limits \tan^{-1} x \, dx = x \tan^{-1} x - \int\limits \frac{x}{1+x^2} \, dx$

To evaluate $\int\limits \frac{x}{1+x^2} \, dx$, we use the substitution $w = 1+x^2$, so $dw = 2x \, dx$, which means $x \, dx = \frac{1}{2} dw$.

$\int\limits \frac{x}{1+x^2} \, dx = \int\limits \frac{1}{w} \cdot \frac{1}{2} dw = \frac{1}{2} \int\limits \frac{1}{w} \, dw = \frac{1}{2} \log|w| + C' = \frac{1}{2} \log(1+x^2) + C'$ (since $1+x^2 > 0$).

Substitute this back into the integral of $\tan^{-1} x$:

$\int\limits \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \log(1+x^2)$ (excluding constant here)

Now, multiply by 2 to get the result for the original integral:

$\int\limits \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \, dx = 2 \int\limits \tan^{-1} x \, dx = 2 \left(x \tan^{-1} x - \frac{1}{2} \log(1+x^2)\right) + C$

= $2x \tan^{-1} x - \log(1+x^2) + C$

where $C$ is the constant of integration.

Final Answer:

The integral of $\sin^{-1} \left( \frac{2x}{1 + x^2} \right)$ is $2x \tan^{-1} x - \log(1+x^2) + C$ (assuming $|x| \leq 1$).

$\int\limits \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \, dx = 2x \tan^{-1} x - \log(1+x^2) + C$

Note: If $|x| > 1$, the identity $\sin^{-1}(\sin(2\theta)) = 2\theta$ does not hold directly. The integrand would be piecewise. However, for typical problems at this level, the assumption $|x| \leq 1$ (or the principal branch where $2\theta$ falls into $[-\pi/2, \pi/2]$) is usually implied.

We need to evaluate the integral $\int\limits x^2 e^{x^3} \, dx$.