Integration by substitution Questions in English

(A) Let $x+a = t$. Then $dx = dt$ and $x = t-a$.
Substituting these into the integral:
$\int \frac{\sin x}{\sin (x+a)} dx = \int \frac{\sin (t-a)}{\sin t} dt$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$= \int \frac{\sin t \cos a - \cos t \sin a}{\sin t} dt$
$= \int (\cos a - \cot t \sin a) dt$
$= \cos a \int dt - \sin a \int \cot t dt$
$= t \cos a - \sin a \log |\sin t| + C$
Substituting $t = x+a$ back:
$= (x+a) \cos a - \sin a \log |\sin (x+a)| + C$
$= x \cos a + a \cos a - \sin a \log |\sin (x+a)| + C$
Since $a \cos a$ is a constant,we can absorb it into the arbitrary constant $C$:
$= x \cos a - \sin a \log |\sin (x+a)| + C$

(A) Let $1 + x^2 = t$.
Then,differentiating both sides with respect to $x$,we get $2x \, dx = dt$.
Substituting these into the integral:
$\int \frac{2x}{1 + x^2} \, dx = \int \frac{1}{t} \, dt$.
Integrating $\frac{1}{t}$ with respect to $t$ gives $\log |t| + C$.
Substituting back $t = 1 + x^2$,we get $\log |1 + x^2| + C$.
Since $1 + x^2$ is always positive for all real $x$,we can write this as $\log (1 + x^2) + C$,where $C$ is an arbitrary constant.

(A) Let $\log x = t$.
Then,differentiating with respect to $x$,we get $\frac{1}{x} dx = dt$.
Substituting these into the integral,we have:
$\int \frac{(\log x)^{2}}{x} dx = \int t^{2} dt$.
Using the power rule for integration $\int x^{n} dx = \frac{x^{n+1}}{n+1} + C$,we get:
$\int t^{2} dt = \frac{t^{3}}{3} + C$.
Substituting $t = \log x$ back into the expression,we obtain:
$\frac{(\log x)^{3}}{3} + C$,where $C$ is an arbitrary constant.

(N/A) The given function can be rewritten as:
$\frac{1}{x+x \log x} = \frac{1}{x(1+\log x)}$
Let $1+\log x = t$.
Then,differentiating both sides with respect to $x$,we get:
$\frac{1}{x} dx = dt$
Substituting these into the integral:
$\int \frac{1}{x(1+\log x)} dx = \int \frac{1}{t} dt$
Integrating with respect to $t$:
$= \log |t| + C$
Substituting back $t = 1 + \log x$:
$= \log |1 + \log x| + C$
where $C$ is an arbitrary constant.

(A) Let $I = \int \sin x \cdot \sin (\cos x) \, dx$.
Substitute $\cos x = t$.
Differentiating both sides with respect to $x$,we get $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Substituting these into the integral,we get:
$I = \int \sin (t) \cdot (-dt) = -\int \sin t \, dt$.
The integral of $\sin t$ is $-\cos t$.
Therefore,$I = -(-\cos t) + C = \cos t + C$.
Substituting back $t = \cos x$,we get the final result:
$I = \cos (\cos x) + C$,where $C$ is an arbitrary constant.

Let $ax+b = t$.
Differentiating both sides with respect to $x$,we get $a \, dx = dt$,which implies $dx = \frac{1}{a} \, dt$.
Substituting these into the integral:
$\int \sqrt{ax+b} \, dx = \int t^{1/2} \cdot \frac{1}{a} \, dt$
$= \frac{1}{a} \int t^{1/2} \, dt$
$= \frac{1}{a} \left( \frac{t^{1/2+1}}{1/2+1} \right) + C$
$= \frac{1}{a} \left( \frac{t^{3/2}}{3/2} \right) + C$
$= \frac{2}{3a} t^{3/2} + C$
Substituting $t = ax+b$ back,we get:
$= \frac{2}{3a} (ax+b)^{3/2} + C$,where $C$ is an arbitrary constant.

Let $x+2=t$.
Then $x=t-2$ and $dx=dt$.
Substituting these into the integral:
$\int x \sqrt{x+2} \, dx = \int (t-2) \sqrt{t} \, dt$
$= \int (t \cdot t^{1/2} - 2t^{1/2}) \, dt$
$= \int (t^{3/2} - 2t^{1/2}) \, dt$
$= \int t^{3/2} \, dt - 2 \int t^{1/2} \, dt$
$= \frac{t^{5/2}}{5/2} - 2 \left( \frac{t^{3/2}}{3/2} \right) + C$
$= \frac{2}{5} t^{5/2} - \frac{4}{3} t^{3/2} + C$
Substituting $t = x+2$ back:
$= \frac{2}{5} (x+2)^{5/2} - \frac{4}{3} (x+2)^{3/2} + C$,where $C$ is the constant of integration.

(N/A) Let $1+2 x^{2} = t$.
Then,differentiating both sides with respect to $x$,we get $4x \, dx = dt$,which implies $x \, dx = \frac{dt}{4}$.
Substituting these into the integral:
$\int x \sqrt{1+2 x^{2}} \, dx = \int \sqrt{t} \cdot \frac{dt}{4} = \frac{1}{4} \int t^{1/2} \, dt$.
Using the power rule $\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$:
$= \frac{1}{4} \left( \frac{t^{3/2}}{3/2} \right) + C = \frac{1}{4} \cdot \frac{2}{3} t^{3/2} + C = \frac{1}{6} t^{3/2} + C$.
Substituting back $t = 1+2x^2$:
$= \frac{1}{6} (1+2x^2)^{3/2} + C$,where $C$ is an arbitrary constant.

(A) Let $I = \int (4x+2)\sqrt{x^{2}+x+1} \, dx$.
Substitute $t = x^{2}+x+1$.
Then,$dt = (2x+1) \, dx$,which implies $2 \, dt = (4x+2) \, dx$.
Substituting these into the integral:
$I = \int 2\sqrt{t} \, dt = 2 \int t^{\frac{1}{2}} \, dt$.
Using the power rule $\int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C$:
$I = 2 \left( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right) + C = 2 \times \frac{2}{3} t^{\frac{3}{2}} + C = \frac{4}{3} t^{\frac{3}{2}} + C$.
Substituting back $t = x^{2}+x+1$:
$I = \frac{4}{3}(x^{2}+x+1)^{\frac{3}{2}} + C$,where $C$ is an arbitrary constant.

(A) The given integral is $I = \int \frac{1}{x-\sqrt{x}} dx$.
We can rewrite the denominator as $\sqrt{x}(\sqrt{x}-1)$.
So,$I = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} dx$.
Let $t = \sqrt{x}-1$.
Then,$dt = \frac{1}{2\sqrt{x}} dx$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$.
Substituting these into the integral,we get $I = \int \frac{2}{t} dt$.
Integrating with respect to $t$,we obtain $I = 2 \log |t| + C$.
Substituting back $t = \sqrt{x}-1$,we get $I = 2 \log |\sqrt{x}-1| + C$,where $C$ is an arbitrary constant.

Let $x+4 = t$.
Then $dx = dt$ and $x = t-4$.
Substituting these into the integral:
$\int \frac{x}{\sqrt{x+4}} dx = \int \frac{t-4}{\sqrt{t}} dt$
$= \int \left( \frac{t}{\sqrt{t}} - \frac{4}{\sqrt{t}} \right) dt$
$= \int (t^{1/2} - 4t^{-1/2}) dt$
$= \frac{t^{3/2}}{3/2} - 4 \left( \frac{t^{1/2}}{1/2} \right) + C$
$= \frac{2}{3} t^{3/2} - 8 t^{1/2} + C$
$= \frac{2}{3} t^{1/2} (t - 12) + C$
Substituting back $t = x+4$:
$= \frac{2}{3} \sqrt{x+4} (x+4-12) + C$
$= \frac{2}{3} \sqrt{x+4} (x-8) + C$,where $C$ is an arbitrary constant.

Let $x^{3}-1=t$.
Then,$3x^{2}dx = dt$,which implies $x^{2}dx = \frac{dt}{3}$.
We can rewrite the integral as:
$\int \left(x^{3}-1\right)^{\frac{1}{3}} x^{5} dx = \int \left(x^{3}-1\right)^{\frac{1}{3}} x^{3} \cdot x^{2} dx$.
Substituting $x^{3} = t+1$ and $x^{2}dx = \frac{dt}{3}$,we get:
$= \int t^{\frac{1}{3}}(t+1) \frac{dt}{3} = \frac{1}{3} \int \left(t^{\frac{4}{3}} + t^{\frac{1}{3}}\right) dt$.
Integrating term by term:
$= \frac{1}{3} \left[ \frac{t^{\frac{7}{3}}}{\frac{7}{3}} + \frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right] + C = \frac{1}{3} \left[ \frac{3}{7} t^{\frac{7}{3}} + \frac{3}{4} t^{\frac{4}{3}} \right] + C$.
$= \frac{1}{7} t^{\frac{7}{3}} + \frac{1}{4} t^{\frac{4}{3}} + C$.
Substituting back $t = x^{3}-1$:
$= \frac{1}{7} \left(x^{3}-1\right)^{\frac{7}{3}} + \frac{1}{4} \left(x^{3}-1\right)^{\frac{4}{3}} + C$,where $C$ is an arbitrary constant.

Let $2+3x^{3} = t$.
Then,differentiating both sides with respect to $x$,we get $9x^{2} dx = dt$,which implies $x^{2} dx = \frac{1}{9} dt$.
Substituting these into the integral:
$\int \frac{x^{2}}{(2+3x^{3})^{3}} dx = \int \frac{1}{9t^{3}} dt = \frac{1}{9} \int t^{-3} dt$.
Using the power rule $\int t^{n} dt = \frac{t^{n+1}}{n+1} + C$:
$= \frac{1}{9} \left( \frac{t^{-2}}{-2} \right) + C = -\frac{1}{18t^{2}} + C$.
Substituting $t = 2+3x^{3}$ back into the expression:
$= -\frac{1}{18(2+3x^{3})^{2}} + C$,where $C$ is an arbitrary constant.

(N/A) Let $\log x = t$.
Then,differentiating both sides with respect to $x$,we get $\frac{1}{x} dx = dt$.
Substituting these into the integral,we have:
$\int \frac{1}{x(\log x)^{m}} dx = \int \frac{1}{t^{m}} dt$.
This can be written as $\int t^{-m} dt$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$),we get:
$\frac{t^{-m+1}}{-m+1} + C = \frac{t^{1-m}}{1-m} + C$.
Substituting back $t = \log x$,the final result is:
$\frac{(\log x)^{1-m}}{1-m} + C$,where $C$ is an arbitrary constant.

(N/A) Let $I = \int \frac{x}{9-4x^{2}} dx$.
Substitute $t = 9-4x^{2}$.
Then,$dt = -8x dx$,which implies $x dx = -\frac{1}{8} dt$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t} \left(-\frac{1}{8}\right) dt$
$I = -\frac{1}{8} \int \frac{1}{t} dt$
$I = -\frac{1}{8} \log |t| + C$
Substituting back $t = 9-4x^{2}$,we get:
$I = -\frac{1}{8} \log |9-4x^{2}| + C$,where $C$ is an arbitrary constant.

(A) To integrate the function $I = \int \frac{x}{e^{x^{2}}} dx$,we use the method of substitution.
Let $x^{2} = t$.
Differentiating both sides with respect to $x$,we get $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \int \frac{1}{e^{t}} \cdot \frac{1}{2} dt$
$I = \frac{1}{2} \int e^{-t} dt$
Integrating $e^{-t}$ with respect to $t$ gives $-e^{-t}$:
$I = \frac{1}{2} (-e^{-t}) + C$
$I = -\frac{1}{2} e^{-t} + C$
Substituting back $t = x^{2}$:
$I = -\frac{1}{2} e^{-x^{2}} + C = -\frac{1}{2e^{x^{2}}} + C$,where $C$ is an arbitrary constant.

(N/A) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^{2}} dx$.
Substitute $\tan ^{-1} x = t$.
Differentiating both sides with respect to $x$,we get $\frac{1}{1+x^{2}} dx = dt$.
Substituting these into the integral,we get $\int e^{t} dt$.
The integral of $e^{t}$ is $e^{t} + C$.
Substituting back $t = \tan ^{-1} x$,we get the final result as $e^{\tan ^{-1} x} + C$,where $C$ is an arbitrary constant.

(N/A) To integrate the function $I = \int \frac{e^{2x}-1}{e^{2x}+1} dx$,we first divide the numerator and the denominator by $e^{x}$:
$I = \int \frac{\frac{e^{2x}-1}{e^{x}}}{\frac{e^{2x}+1}{e^{x}}} dx = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx$
Now,let $t = e^{x}+e^{-x}$.
Differentiating both sides with respect to $x$,we get $dt = (e^{x}-e^{-x}) dx$.
Substituting these into the integral,we have:
$I = \int \frac{dt}{t} = \log |t| + C$
Substituting back the value of $t$,we get:
$I = \log |e^{x}+e^{-x}| + C$,where $C$ is an arbitrary constant.

Let $I = \int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} dx$.
Substitute $t = e^{2 x}+e^{-2 x}$.
Differentiating both sides with respect to $x$,we get:
$dt = (2e^{2 x} - 2e^{-2 x}) dx = 2(e^{2 x} - e^{-2 x}) dx$.
Therefore,$(e^{2 x} - e^{-2 x}) dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{1}{t} dt$.
Integrating,we get:
$I = \frac{1}{2} \log |t| + C$.
Substituting back $t = e^{2 x}+e^{-2 x}$:
$I = \frac{1}{2} \log |e^{2 x}+e^{-2 x}| + C$,where $C$ is an arbitrary constant.

(A) Let $7-4x = t$.
Then,differentiating both sides with respect to $x$,we get $-4 dx = dt$,which implies $dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$\int \sec^{2}(7-4x) dx = \int \sec^{2}(t) \left(-\frac{1}{4}\right) dt$
$= -\frac{1}{4} \int \sec^{2}(t) dt$
$= -\frac{1}{4} \tan(t) + C$
$= -\frac{1}{4} \tan(7-4x) + C$,where $C$ is an arbitrary constant.

Let $t = \sin^{-1} x$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sqrt{1-x^2}}$,which implies $dt = \frac{1}{\sqrt{1-x^2}} \, dx$.
Substituting these into the integral,we have:
$\int \frac{\sin^{-1} x}{\sqrt{1-x^2}} \, dx = \int t \, dt$.
Integrating $t$ with respect to $t$ gives $\frac{t^2}{2} + C$.
Substituting back $t = \sin^{-1} x$,the final result is $\frac{(\sin^{-1} x)^2}{2} + C$,where $C$ is an arbitrary constant.

Let $I = \int \frac{2 \cos x - 3 \sin x}{6 \cos x + 4 \sin x} dx$.
We can rewrite the denominator as $2(3 \cos x + 2 \sin x)$.
So,$I = \int \frac{2 \cos x - 3 \sin x}{2(3 \cos x + 2 \sin x)} dx$.
Let $t = 3 \cos x + 2 \sin x$.
Differentiating with respect to $x$,we get $dt = (-3 \sin x + 2 \cos x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{2t} = \frac{1}{2} \int \frac{1}{t} dt$.
Integrating,we get $I = \frac{1}{2} \log |t| + C$.
Substituting back $t = 3 \cos x + 2 \sin x$,we get $I = \frac{1}{2} \log |3 \cos x + 2 \sin x| + C$,where $C$ is an arbitrary constant.

(A) The given integral is $I = \int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} dx$.
We know that $\frac{1}{\cos ^{2} x} = \sec ^{2} x$,so the integral becomes $I = \int \frac{\sec ^{2} x}{(1-\tan x)^{2}} dx$.
Let $u = 1 - \tan x$.
Then,differentiating with respect to $x$,we get $du = -\sec ^{2} x dx$,which implies $\sec ^{2} x dx = -du$.
Substituting these into the integral,we get $I = \int \frac{-du}{u^{2}} = -\int u^{-2} du$.
Using the power rule for integration,$\int u^{n} du = \frac{u^{n+1}}{n+1}$,we get $I = -\left( \frac{u^{-1}}{-1} \right) + C = \frac{1}{u} + C$.
Substituting back $u = 1 - \tan x$,we get $I = \frac{1}{1-\tan x} + C$,where $C$ is an arbitrary constant.

(A) Let $\sqrt{x} = t$.
Then,differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{1}{\sqrt{x}} dx = 2 dt$.
Substituting these into the integral:
$\int \frac{\cos \sqrt{x}}{\sqrt{x}} dx = \int \cos(t) \cdot (2 dt) = 2 \int \cos(t) dt$.
Integrating $\cos(t)$ gives $\sin(t)$,so we have $2 \sin(t) + C$.
Substituting back $t = \sqrt{x}$,the final result is $2 \sin \sqrt{x} + C$,where $C$ is an arbitrary constant.

(A) Let $I = \int \sqrt{\sin 2x} \cos 2x \, dx$.
Substitute $\sin 2x = t$.
Differentiating both sides with respect to $x$,we get:
$2 \cos 2x \, dx = dt$
$\cos 2x \, dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \int \sqrt{t} \cdot \frac{1}{2} dt$
$I = \frac{1}{2} \int t^{\frac{1}{2}} dt$
Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
$I = \frac{1}{2} \left( \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} \right) + C$
$I = \frac{1}{2} \left( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right) + C$
$I = \frac{1}{2} \cdot \frac{2}{3} t^{\frac{3}{2}} + C$
$I = \frac{1}{3} t^{\frac{3}{2}} + C$
Substituting $t = \sin 2x$ back:
$I = \frac{1}{3} (\sin 2x)^{\frac{3}{2}} + C$,where $C$ is the constant of integration.

(A) Let $1+\sin x = t$.
Differentiating both sides with respect to $x$,we get $\cos x \, dx = dt$.
Substituting these into the integral:
$\int \frac{\cos x}{\sqrt{1+\sin x}} \, dx = \int \frac{dt}{\sqrt{t}} = \int t^{-\frac{1}{2}} \, dt$.
Using the power rule $\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$:
$= \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + C = 2\sqrt{t} + C$.
Substituting back $t = 1+\sin x$:
$= 2\sqrt{1+\sin x} + C$,where $C$ is an arbitrary constant.

(N/A) Let $I = \int \cot x \log \sin x \, dx$.
Substitute $t = \log \sin x$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Therefore,$dt = \cot x \, dx$.
Substituting these into the integral,we get $I = \int t \, dt$.
Integrating with respect to $t$,we get $I = \frac{t^2}{2} + C$.
Substituting back $t = \log \sin x$,we get $I = \frac{1}{2}(\log \sin x)^2 + C$,where $C$ is an arbitrary constant.

(A) Let $I = \int \frac{\sin x}{1+\cos x} dx$.
Substitute $t = 1 + \cos x$.
Differentiating with respect to $x$,we get $dt = -\sin x dx$,which implies $\sin x dx = -dt$.
Substituting these into the integral,we get $I = \int -\frac{dt}{t}$.
Integrating,we obtain $I = -\log |t| + C$.
Substituting back $t = 1 + \cos x$,we get $I = -\log |1 + \cos x| + C$,where $C$ is an arbitrary constant.

Let $1+\cos x = t$.
Differentiating both sides with respect to $x$,we get $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Substituting these into the integral:
$\int \frac{\sin x}{(1+\cos x)^{2}} \, dx = \int -\frac{dt}{t^{2}}$.
$= -\int t^{-2} \, dt$.
$= -\left( \frac{t^{-1}}{-1} \right) + C$.
$= \frac{1}{t} + C$.
Substituting back $t = 1+\cos x$,we get:
$= \frac{1}{1+\cos x} + C$,where $C$ is an arbitrary constant.

(A) Let $I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$.
Multiply the numerator and denominator by $\cos x$:
$I = \int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cos x \cdot \cos x} dx = \int \frac{\sqrt{\tan x}}{\tan x \cos^2 x} dx$.
Since $\frac{1}{\cos^2 x} = \sec^2 x$,we have:
$I = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$.
Let $t = \tan x$,then $dt = \sec^2 x dx$.
Substituting these into the integral:
$I = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt$.
Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + C$:
$I = \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$.
Substituting back $t = \tan x$:
$I = 2\sqrt{\tan x} + C$,where $C$ is an arbitrary constant.

(A) Let $1+\log x = t$.
Differentiating both sides with respect to $x$,we get $\frac{1}{x} dx = dt$.
Substituting these into the integral,we have:
$\int \frac{(1+\log x)^{2}}{x} dx = \int t^{2} dt$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$,we get:
$\int t^{2} dt = \frac{t^{3}}{3} + C$.
Substituting $t = 1+\log x$ back into the expression,we get:
$\frac{(1+\log x)^{3}}{3} + C$,where $C$ is an arbitrary constant.

(N/A) The given function can be rewritten as:
$\frac{(x+1)(x+\log x)^{2}}{x} = \left(1+\frac{1}{x}\right)(x+\log x)^{2}$.
Let $t = x + \log x$.
Then,differentiating with respect to $x$,we get $dt = (1 + \frac{1}{x}) dx$.
Substituting these into the integral,we have:
$\int (x + \log x)^{2} (1 + \frac{1}{x}) dx = \int t^{2} dt$.
Integrating $t^{2}$ with respect to $t$ gives $\frac{t^{3}}{3} + C$.
Substituting back $t = x + \log x$,the final result is $\frac{1}{3}(x + \log x)^{3} + C$,where $C$ is an arbitrary constant.

Let $x^{4} = t$.
Then,$4x^{3} dx = dt$,which implies $x^{3} dx = \frac{1}{4} dt$.
Substituting this into the integral,we get:
$\int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}} dx = \frac{1}{4} \int \frac{\sin \left(\tan ^{-1} t\right)}{1+t^{2}} dt$.
Now,let $\tan ^{-1} t = u$.
Then,$\frac{1}{1+t^{2}} dt = du$.
Substituting $u$ into the integral:
$\frac{1}{4} \int \sin u du = \frac{1}{4} (-\cos u) + C = -\frac{1}{4} \cos u + C$.
Substituting back $u = \tan ^{-1} t$ and $t = x^{4}$:
$= -\frac{1}{4} \cos \left(\tan ^{-1} x^{4}\right) + C$,where $C$ is the constant of integration.

(D) Let $t = x^{10} + 10^{x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(10^{x})$
$\frac{dt}{dx} = 10x^{9} + 10^{x} \log_{e} 10$
Therefore,$dt = (10x^{9} + 10^{x} \log_{e} 10) dx$.
Substituting these into the integral,we have:
$\int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x = \int \frac{1}{t} dt$
$= \log |t| + C$
$= \log |x^{10} + 10^{x}| + C$
Since $x^{10} + 10^{x} > 0$ for all real $x$,we can write this as $\log (x^{10} + 10^{x}) + C$.
Hence,the correct option is $D$.

Let $I = \int \sin ^{3}(2 x+1) dx$.
We can write $\sin ^{3}(2 x+1) = \sin ^{2}(2 x+1) \cdot \sin (2 x+1)$.
Using the identity $\sin ^{2}\theta = 1 - \cos ^{2}\theta$,we get:
$I = \int (1 - \cos ^{2}(2 x+1)) \sin (2 x+1) dx$.
Let $t = \cos (2 x+1)$.
Then $dt = -2 \sin (2 x+1) dx$,which implies $\sin (2 x+1) dx = -\frac{dt}{2}$.
Substituting these into the integral:
$I = \int (1 - t^{2}) \left(-\frac{dt}{2}\right) = -\frac{1}{2} \int (1 - t^{2}) dt$.
Integrating with respect to $t$:
$I = -\frac{1}{2} \left(t - \frac{t^{3}}{3}\right) + C$.
Substituting $t = \cos (2 x+1)$ back:
$I = -\frac{1}{2} \cos (2 x+1) + \frac{1}{6} \cos ^{3}(2 x+1) + C$,where $C$ is an arbitrary constant.

Let $I = \int \sin ^{3} x \cos ^{3} x \, dx$.
We can rewrite the integral as:
$I = \int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \, dx$.
Using the identity $\sin ^{2} x = 1 - \cos ^{2} x$:
$I = \int \cos ^{3} x (1 - \cos ^{2} x) \sin x \, dx$.
Let $\cos x = t$. Then,$-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Substituting these into the integral:
$I = -\int t^{3} (1 - t^{2}) \, dt$.
$I = -\int (t^{3} - t^{5}) \, dt$.
Integrating with respect to $t$:
$I = -\left( \frac{t^{4}}{4} - \frac{t^{6}}{6} \right) + C$.
$I = \frac{t^{6}}{6} - \frac{t^{4}}{4} + C$.
Substituting back $t = \cos x$:
$I = \frac{\cos ^{6} x}{6} - \frac{\cos ^{4} x}{4} + C$,where $C$ is an arbitrary constant.

(N/A) We have the integral $I = \int \frac{\cos x-\sin x}{1+\sin 2 x} dx$.
Using the identities $\sin^2 x + \cos^2 x = 1$ and $\sin 2x = 2 \sin x \cos x$,the denominator becomes:
$1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$.
Thus,the integral is $I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{t^2} = \int t^{-2} dt$.
Integrating with respect to $t$:
$I = \frac{t^{-2+1}}{-2+1} + C = -t^{-1} + C = -\frac{1}{t} + C$.
Substituting back $t = \sin x + \cos x$:
$I = -\frac{1}{\sin x + \cos x} + C$,where $C$ is an arbitrary constant.

(D) We have the integral $I = \int \frac{\cos 2x}{(\cos x + \sin x)^2} dx$.
Using the trigonometric identity $\cos 2x = \cos^2 x - \sin^2 x$ and the expansion $(\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2 \sin x \cos x = 1 + \sin 2x$,we get:
$I = \int \frac{\cos^2 x - \sin^2 x}{1 + \sin 2x} dx = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} dx$
$I = \int \frac{\cos x - \sin x}{\cos x + \sin x} dx$
Let $u = \cos x + \sin x$. Then $du = (-\sin x + \cos x) dx$.
Substituting these into the integral:
$I = \int \frac{1}{u} du = \log |u| + C$
$I = \log |\cos x + \sin x| + C$,where $C$ is an arbitrary constant.

(A) Let $I = \int \frac{d x}{e^{x}+e^{-x}}$.
Multiply the numerator and denominator by $e^{x}$:
$I = \int \frac{e^{x}}{e^{2 x}+1} d x$.
Let $e^{x} = t$. Then $e^{x} d x = d t$.
Substituting these into the integral:
$I = \int \frac{d t}{t^{2}+1}$.
Using the standard integral formula $\int \frac{d x}{x^{2}+1} = \tan ^{-1}(x) + C$:
$I = \tan ^{-1}(t) + C$.
Substituting $t = e^{x}$ back:
$I = \tan ^{-1}\left(e^{x}\right) + C$.
Thus,the correct option is $A$.

(B) Let $I = \int \frac{\cos 2x}{(\sin x + \cos x)^2} dx$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have:
$I = \int \frac{\cos^2 x - \sin^2 x}{(\sin x + \cos x)^2} dx$.
Factorizing the numerator as a difference of squares:
$I = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} dx$.
Simplifying the expression:
$I = \int \frac{\cos x - \sin x}{\sin x + \cos x} dx$.
Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral:
$I = \int \frac{1}{t} dt = \log |t| + C$.
Substituting back $t = \sin x + \cos x$:
$I = \log |\sin x + \cos x| + C$.
Thus,the correct option is $B$.

(D) Let $I = \int \frac{e^{x}(1+x)}{\cos ^{2}(x e^{x})} d x$.
Substitute $t = x e^{x}$.
Differentiating both sides with respect to $x$,we get:
$dt = (e^{x} + x e^{x}) dx = e^{x}(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{\cos ^{2} t} = \int \sec ^{2} t dt$.
Integrating $\sec ^{2} t$,we get:
$I = \tan t + C$.
Substituting back $t = x e^{x}$:
$I = \tan (x e^{x}) + C$.
Thus,the correct option is $D$.

(A) To evaluate the integral $\int \frac{dx}{\sqrt{2x-x^2}}$,we first complete the square in the denominator:
$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x-1)^2$
So,the integral becomes $\int \frac{dx}{\sqrt{1-(x-1)^2}}$
Let $t = x-1$,then $dt = dx$
Substituting these into the integral,we get $\int \frac{dt}{\sqrt{1-t^2}}$
Using the standard integral formula $\int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1}(t) + C$
Substituting back $t = x-1$,we get $\sin^{-1}(x-1) + C$

(A) Let $x^{3} = t$.
Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$.
Substituting these into the integral,we have:
$\int \frac{3x^{2}}{x^{6}+1} dx = \int \frac{dt}{t^{2}+1}$.
Using the standard integral formula $\int \frac{1}{x^{2}+1} dx = \tan^{-1}(x) + C$,we get:
$= \tan^{-1}(t) + C$.
Substituting back $t = x^{3}$,the final result is:
$= \tan^{-1}(x^{3}) + C$,where $C$ is an arbitrary constant.

Let $\sqrt{2}x^2 = t$.
Then,differentiating both sides with respect to $x$,we get $2\sqrt{2}x dx = dt$,which implies $x dx = \frac{dt}{2\sqrt{2}}$.
Substituting these into the integral:
$\int \frac{3x}{1+2x^4} dx = 3 \int \frac{1}{1+t^2} \cdot \frac{dt}{2\sqrt{2}}$
$= \frac{3}{2\sqrt{2}} \int \frac{1}{1+t^2} dt$
$= \frac{3}{2\sqrt{2}} \tan^{-1}(t) + C$
$= \frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2}x^2) + C$,where $C$ is the constant of integration.

Let $x^{3} = t$.
Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$,which implies $x^{2} dx = \frac{1}{3} dt$.
Substituting these into the integral:
$\int \frac{x^{2}}{1-x^{6}} dx = \int \frac{1}{1-(x^{3})^{2}} (x^{2} dx) = \frac{1}{3} \int \frac{dt}{1-t^{2}}$.
Using the standard integral formula $\int \frac{1}{a^{2}-x^{2}} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,where $a=1$:
$= \frac{1}{3} \left[ \frac{1}{2(1)} \log \left| \frac{1+t}{1-t} \right| \right] + C$
$= \frac{1}{6} \log \left| \frac{1+x^{3}}{1-x^{3}} \right| + C$,where $C$ is an arbitrary constant.

(A) Let $x^{3} = t$.
Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$,which implies $x^{2} dx = \frac{1}{3} dt$.
Substituting these into the integral:
$\int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} dx = \int \frac{1}{\sqrt{(x^{3})^{2} + (a^{3})^{2}}} \cdot \frac{1}{3} dt$
$= \frac{1}{3} \int \frac{dt}{\sqrt{t^{2} + (a^{3})^{2}}}$
Using the standard integral formula $\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \log |x + \sqrt{x^{2} + a^{2}}| + C$,we get:
$= \frac{1}{3} \log |t + \sqrt{t^{2} + (a^{3})^{2}}| + C$
Substituting $t = x^{3}$ back into the expression:
$= \frac{1}{3} \log |x^{3} + \sqrt{x^{6} + a^{6}}| + C$,where $C$ is an arbitrary constant.

(N/A) Let $\tan x = t$.
Then,$\sec ^{2} x \, dx = dt$.
Substituting these into the integral,we get:
$\int \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} \, dx = \int \frac{dt}{\sqrt{t^{2} + 2^{2}}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \log |x + \sqrt{x^{2} + a^{2}}| + C$,we have:
$= \log |t + \sqrt{t^{2} + 4}| + C$.
Substituting back $t = \tan x$,the final result is:
$= \log |\tan x + \sqrt{\tan^{2} x + 4}| + C$,where $C$ is an arbitrary constant.

(N/A) Let $2x^{2}+x-3 = t$.
Then,differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(2x^{2}+x-3) = \frac{dt}{dx}$
$(4x+1) dx = dt$.
Substituting these into the integral:
$\int \frac{4x+1}{\sqrt{2x^{2}+x-3}} dx = \int \frac{1}{\sqrt{t}} dt$.
This is equivalent to $\int t^{-1/2} dt$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
$= \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$.
Substituting the value of $t$ back:
$= 2\sqrt{2x^{2}+x-3} + C$,where $C$ is an arbitrary constant.

(N/A) Let $y = \sin \phi$.
Then $dy = \cos \phi \, d\phi$.
Substituting these into the integral,we get:
$\int \frac{(3y-2) dy}{5 - (1 - y^2) - 4y} = \int \frac{3y-2}{y^2 - 4y + 4} dy = \int \frac{3y-2}{(y-2)^2} dy$.
Using partial fractions,let $\frac{3y-2}{(y-2)^2} = \frac{A}{y-2} + \frac{B}{(y-2)^2}$.
Then $3y-2 = A(y-2) + B$.
Comparing coefficients,$A = 3$ and $-2A + B = -2 \implies B = 4$.
Thus,the integral becomes $\int \left( \frac{3}{y-2} + \frac{4}{(y-2)^2} \right) dy$.
$= 3 \ln |y-2| - \frac{4}{y-2} + C$.
Substituting $y = \sin \phi$ back,we get $3 \ln |\sin \phi - 2| - \frac{4}{\sin \phi - 2} + C$.
Since $\sin \phi - 2$ is always negative,$|\sin \phi - 2| = 2 - \sin \phi$.
$= 3 \ln (2 - \sin \phi) + \frac{4}{2 - \sin \phi} + C$.

(C) Let $I = \int x^{2} e^{x^{3}} d x$.
Substitute $t = x^{3}$,then differentiate both sides with respect to $x$:
$dt = 3x^{2} dx$,which implies $x^{2} dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int e^{t} \cdot \frac{1}{3} dt$
$I = \frac{1}{3} \int e^{t} dt$
Integrating $e^{t}$ gives $e^{t}$:
$I = \frac{1}{3} e^{t} + C$
Substituting back $t = x^{3}$:
$I = \frac{1}{3} e^{x^{3}} + C$.
Thus,the correct option is $C$.