Integration by substitution Questions in English

(B) To evaluate the integral $I = \int \frac{\sin x}{\sin (x - \alpha )} dx$,we rewrite the numerator as $\sin(x - \alpha + \alpha)$.
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get:
$I = \int \frac{\sin(x - \alpha) \cos \alpha + \cos(x - \alpha) \sin \alpha}{\sin(x - \alpha)} dx$
$I = \int \left( \cos \alpha + \sin \alpha \cdot \cot(x - \alpha) \right) dx$
$I = \int \cos \alpha \, dx + \sin \alpha \int \cot(x - \alpha) \, dx$
Since $\int \cos \alpha \, dx = x \cos \alpha$ and $\int \cot(x - \alpha) \, dx = \log |\sin(x - \alpha)|$,we have:
$I = x \cos \alpha + \sin \alpha \log |\sin(x - \alpha)| + c$.

(A) Given that $\int {f(x)\,dx = f(x)} $.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx} \left( \int {f(x)\,dx} \right) = \frac{d}{dx} f(x)$
$f(x) = f'(x)$.
Now,we need to evaluate $\int {{{\left[ {f(x)} \right]}^2}\,dx}$.
Since $f(x) = f'(x)$,we can write the integral as:
$\int {{{\left[ {f(x)} \right]}^2}\,dx} = \int {f(x) \cdot f'(x)\,dx}$.
Let $f(x) = u$,then $f'(x)\,dx = du$.
The integral becomes $\int {u\,du} = \frac{u^2}{2} + C$.
Substituting $u = f(x)$ back,we get $\frac{1}{2}{\left[ {f(x)} \right]^2} + C$.
Thus,the correct option is $A$.

(B) To evaluate the integral $I = \int \frac{dx}{1 + e^x}$,multiply the numerator and denominator by $e^{-x}$:
$I = \int \frac{e^{-x}}{e^{-x}(1 + e^x)} dx = \int \frac{e^{-x}}{e^{-x} + 1} dx$
Let $t = 1 + e^{-x}$. Then,differentiating with respect to $x$,we get $dt = -e^{-x} dx$,which implies $e^{-x} dx = -dt$.
Substituting these into the integral:
$I = \int \frac{-dt}{t} = -\int \frac{1}{t} dt = -\log|t| + C$
Substituting back $t = 1 + e^{-x}$:
$I = -\log(1 + e^{-x}) + C$
Thus,the correct option is $B$.

(B) Let $I = \int \frac{dx}{e^x + e^{-x}}$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{e^x}{e^{2x} + 1} dx$.
Let $e^x = t$. Then $e^x dx = dt$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + 1}$.
The integral of $\frac{1}{t^2 + 1}$ is $\tan^{-1}(t) + C$.
Thus,$I = \tan^{-1}(e^x) + C$.

(A) Let $I = \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx$.
Substitute $t = e^{\sqrt{x}}$.
Differentiating both sides with respect to $x$,we get:
$dt = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} dx$.
This implies $\frac{e^{\sqrt{x}}}{\sqrt{x}} dx = 2 dt$.
Substituting these into the integral:
$I = \int \cos(t) \cdot 2 dt = 2 \int \cos(t) dt$.
Integrating $\cos(t)$,we get $2 \sin(t) + C$.
Substituting back $t = e^{\sqrt{x}}$,we get $2 \sin(e^{\sqrt{x}}) + C$.

(A) We are given the integral $I = \int \frac{dx}{x + x \log x}$.
First,factor out $x$ from the denominator:
$I = \int \frac{dx}{x(1 + \log x)}$.
Now,use the method of substitution. Let $t = 1 + \log x$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $\frac{dx}{x} = dt$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{t}$.
The integral of $\frac{1}{t}$ with respect to $t$ is $\log|t| + C$.
Substituting back $t = 1 + \log x$,we get:
$I = \log|1 + \log x| + C$.

(B) Let $1 + \log x = t$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(1 + \log x) = \frac{dt}{dx}$.
This implies $\frac{1}{x} dx = dt$.
Substituting these into the integral,we get $\int (1 + \log x) \cdot \frac{1}{x} dx = \int t dt$.
Integrating with respect to $t$,we get $\frac{t^2}{2} + C$.
Substituting back $t = 1 + \log x$,the final result is $\frac{(1 + \log x)^2}{2} + C$.

(A) We have the integral $I = \int \frac{\sec x \, dx}{\sqrt{\cos 2x}}$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we get:
$I = \int \frac{\sec x \, dx}{\sqrt{\cos^2 x - \sin^2 x}}$.
Multiply the numerator and denominator by $\sec x$:
$I = \int \frac{\sec^2 x \, dx}{\sqrt{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}}} = \int \frac{\sec^2 x \, dx}{\sqrt{1 - \tan^2 x}}$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$.
Substituting these into the integral:
$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + C = \sin^{-1}(\tan x) + C$.

(D) To evaluate the integral $I = \int \frac{dx}{x\sqrt{2ax - x^2}}$,we use the substitution $x = \frac{2a}{t}$.
Then,$dx = -\frac{2a}{t^2} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{2a}{t^2} dt}{\frac{2a}{t} \sqrt{2a(\frac{2a}{t}) - (\frac{2a}{t})^2}}$
$I = \int \frac{-\frac{2a}{t^2} dt}{\frac{2a}{t} \sqrt{\frac{4a^2}{t} - \frac{4a^2}{t^2}}}$
$I = \int \frac{-\frac{2a}{t^2} dt}{\frac{2a}{t} \cdot \frac{2a}{t} \sqrt{t - 1}}$
$I = -\int \frac{dt}{t \sqrt{t - 1}}$.
This substitution simplifies the integral into a standard form. Note that the original option $(d)$ in the prompt was $x = 2a \sin^2 t$,which is also a valid substitution,but $x = \frac{2a}{t}$ is the standard substitution for this specific form.

(C) We have the integral $I = \int \frac{x \, dx}{1 - x \cot x}$.
Substitute $\cot x = \frac{\cos x}{\sin x}$:
$I = \int \frac{x \, dx}{1 - x \frac{\cos x}{\sin x}} = \int \frac{x \sin x}{\sin x - x \cos x} \, dx$.
Let $t = \sin x - x \cos x$.
Differentiating with respect to $x$:
$dt = (\cos x - (x(-\sin x) + \cos x)) \, dx = (\cos x + x \sin x - \cos x) \, dx = x \sin x \, dx$.
Substituting into the integral:
$I = \int \frac{dt}{t} = \log |t| + c = \log |\sin x - x \cos x| + c$.

(B) Let $I = \int \frac{\sin 2x}{1 + \sin^2 x} dx$.
Substitute $t = 1 + \sin^2 x$.
Differentiating both sides with respect to $x$,we get $dt = \frac{d}{dx}(1 + \sin^2 x) dx = 2 \sin x \cos x dx = \sin 2x dx$.
Substituting these into the integral,we get $I = \int \frac{1}{t} dt$.
Integrating,we get $I = \log |t| + c$.
Substituting back $t = 1 + \sin^2 x$,we get $I = \log (1 + \sin^2 x) + c$ (since $1 + \sin^2 x > 0$ for all $x$).

(B) Let $I = \int \frac{x^3}{\sqrt{x^2 + 2}} \, dx$.
Substitute $x^2 + 2 = t^2$,which implies $2x \, dx = 2t \, dt$ or $x \, dx = t \, dt$.
Also,$x^2 = t^2 - 2$.
Substituting these into the integral:
$I = \int \frac{(t^2 - 2)}{t} \cdot t \, dt = \int (t^2 - 2) \, dt$.
Integrating with respect to $t$:
$I = \frac{t^3}{3} - 2t + c$.
Substituting $t = \sqrt{x^2 + 2}$ back into the expression:
$I = \frac{1}{3}(x^2 + 2)^{3/2} - 2(x^2 + 2)^{1/2} + c$.

(A) Let $I = \int {\frac{{{x^{e - 1}} + {e^{x - 1}}}}{{{x^e} + {e^x}}}} \,dx$.
Substitute $t = {x^e} + {e^x}$.
Differentiating both sides with respect to $x$,we get $dt = (e \cdot {x^{e - 1}} + {e^x}) \,dx$.
Wait,the derivative of ${e^x}$ is ${e^x}$,not ${e^{x-1}}$. Let us re-evaluate the integral.
Actually,the derivative of ${x^e}$ is $e{x^{e-1}}$ and the derivative of ${e^x}$ is ${e^x}$.
Given the structure,if the integral is $\int \frac{e{x^{e-1}} + {e^x}}{{x^e} + {e^x}} dx$,the answer would be $\log({x^e} + {e^x}) + c$.
However,for the expression $\int \frac{{x^{e-1}} + {e^{x-1}}}{{x^e} + {e^x}} dx$,it does not simplify to a standard logarithmic form directly.
Assuming the intended question was $\int \frac{e{x^{e-1}} + {e^x}}{{x^e} + {e^x}} dx$,the answer is $\log({x^e} + {e^x}) + c$.
If we strictly follow the provided expression,none of the options match.
Given the options,it is highly probable the numerator was meant to be $e{x^{e-1}} + {e^x}$.
Thus,the correct option is $A$.

(C) Let $I = \int \frac{\sin x \, dx}{a^2 + b^2 \cos^2 x}$.
Substitute $t = b \cos x$. Then $dt = -b \sin x \, dx$,which implies $\sin x \, dx = -\frac{dt}{b}$.
The integral becomes $I = \int \frac{-dt/b}{a^2 + t^2} = -\frac{1}{b} \int \frac{dt}{a^2 + t^2}$.
Using the standard integral formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + c$,we get $I = -\frac{1}{b} \cdot \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + c = -\frac{1}{ab} \tan^{-1} \left( \frac{b \cos x}{a} \right) + c$.
Using the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,we have $-\tan^{-1}(x) = \cot^{-1}(x) - \frac{\pi}{2}$.
Thus,$I = \frac{1}{ab} \cot^{-1} \left( \frac{b \cos x}{a} \right) + C$,where $C = c - \frac{\pi}{2ab}$ is a constant.
Therefore,the correct option is $C$.

(B) Let $t = \log(\sec x + \tan x)$.
Then,differentiating with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \sec x$.
Thus,$dt = \sec x \, dx$.
Substituting these into the integral,we get $\int t \, dt$.
Integrating with respect to $t$,we have $\frac{t^2}{2} + c$.
Substituting back the value of $t$,we get $\frac{1}{2}[\log(\sec x + \tan x)]^2 + c$.

(B) Let $I = \int \frac{3x^2}{x^6 + 1} dx$.
Substitute $t = x^3$.
Then,differentiating both sides with respect to $x$,we get $dt = 3x^2 dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t^2 + 1} dt$.
We know that $\int \frac{1}{t^2 + 1} dt = \tan^{-1}(t) + c$.
Substituting $t = x^3$ back,we get $I = \tan^{-1}(x^3) + c$.

(A) Let $I = \int \frac{\cot x}{\log(\sin x)} \, dx$.
Substitute $t = \log(\sin x)$.
Differentiating both sides with respect to $x$,we get:
$dt = \frac{1}{\sin x} \cdot \cos x \, dx = \cot x \, dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t} \, dt$.
Integrating with respect to $t$,we get:
$I = \log|t| + c$.
Substituting back the value of $t$,we get:
$I = \log|\log(\sin x)| + c$.

(C) Let $I = \int \frac{(1 + \log x)^2}{x} \, dx$.
Substitute $t = 1 + \log x$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $\frac{1}{x} \, dx = dt$.
Substituting these into the integral,we get $I = \int t^2 \, dt$.
Integrating $t^2$ with respect to $t$,we get $I = \frac{t^3}{3} + c$.
Substituting back $t = 1 + \log x$,we get $I = \frac{(1 + \log x)^3}{3} + c$.

(B) Let $I = \int \sec^p x \tan x \, dx$.
Substitute $t = \sec x$.
Then,$dt = \sec x \tan x \, dx$,which implies $\tan x \, dx = \frac{dt}{\sec x} = \frac{dt}{t}$.
Substituting these into the integral:
$I = \int t^p \cdot \frac{dt}{t} = \int t^{p-1} \, dt$.
Integrating with respect to $t$:
$I = \frac{t^{(p-1)+1}}{(p-1)+1} + c = \frac{t^p}{p} + c$.
Replacing $t$ with $\sec x$:
$I = \frac{\sec^p x}{p} + c$.

(A) To solve the integral $I = \int \frac{dx}{e^x - 1}$,multiply the numerator and denominator by $e^{-x}$:
$I = \int \frac{e^{-x}}{e^{-x}(e^x - 1)} dx = \int \frac{e^{-x}}{1 - e^{-x}} dx$
Let $t = 1 - e^{-x}$.
Then,the derivative is $dt = e^{-x} dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{t} = \ln|t| + c$
Substituting back $t = 1 - e^{-x}$,we get:
$I = \ln(1 - e^{-x}) + c$.

(C) Let $I = \int x^2 \sec(x^3) \, dx$.
Substitute $t = x^3$.
Then,$dt = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} \, dt$.
Substituting these into the integral:
$I = \int \sec(t) \cdot \frac{1}{3} \, dt$
$I = \frac{1}{3} \int \sec(t) \, dt$
Using the standard integral formula $\int \sec(t) \, dt = \log|\sec(t) + \tan(t)| + C$:
$I = \frac{1}{3} \log|\sec(t) + \tan(t)| + C$
Substituting $t = x^3$ back:
$I = \frac{1}{3} \log|\sec(x^3) + \tan(x^3)| + C$.
Thus,the correct option is $C$.

(B) Let $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} \, dx$.
Divide the numerator and denominator by $\cos^4 x$:
$I = \int \frac{2 \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx = \int \frac{2 \tan x \sec^2 x}{1 + \tan^4 x} \, dx$.
Let $t = \tan^2 x$,then $dt = 2 \tan x \sec^2 x \, dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{1 + t^2} = \tan^{-1}(t) + c$.
Replacing $t$ with $\tan^2 x$,we obtain:
$I = \tan^{-1}(\tan^2 x) + c$.

(B) Let $I = \int \frac{x - 2}{x(2\log x - x)} dx$.
We can rewrite the integrand as:
$I = \int \frac{-(2 - x)}{x(2\log x - x)} dx = - \int \frac{\frac{2}{x} - 1}{2\log x - x} dx$.
Let $t = 2\log x - x$.
Then,differentiating with respect to $x$,we get $dt = (\frac{2}{x} - 1) dx$.
Substituting these into the integral,we get:
$I = - \int \frac{1}{t} dt = - \log |t| + c$.
Substituting $t = 2\log x - x$ back,we get:
$I = - \log |2\log x - x| + c = \log \left( \frac{1}{|2\log x - x|} \right) + c$.

(D) To evaluate the integral $\int x\sqrt{1 + x^2} \, dx$,we use the method of substitution.
Let $t = 1 + x^2$.
Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.
Substituting these into the integral,we get:
$\int \sqrt{t} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{1/2} \, dt$.
Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$,we have:
$\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + c = \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + c = \frac{1}{3} t^{3/2} + c$.
Finally,substituting $t = 1 + x^2$ back,we get $\frac{1}{3}(1 + x^2)^{3/2} + c$.

(A) Let $I = \int \frac{e^x(x + 1)}{\cos^2(x e^x)} dx$.
We can rewrite the integral as $I = \int e^x(x + 1) \sec^2(x e^x) dx$.
Let $t = x e^x$.
Then,differentiating with respect to $x$ using the product rule,we get $dt = (1 \cdot e^x + x \cdot e^x) dx = e^x(1 + x) dx$.
Substituting these into the integral,we get $I = \int \sec^2(t) dt$.
The integral of $\sec^2(t)$ is $\tan(t) + c$.
Substituting back $t = x e^x$,we get $I = \tan(x e^x) + c$.

(B) Let $I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$.
Substitute $\sqrt{x} = t$.
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,we get $I = \int \cos(t) \cdot 2 dt = 2 \int \cos(t) dt$.
Integrating $\cos(t)$,we get $I = 2 \sin(t) + c$.
Substituting back $t = \sqrt{x}$,we get $I = 2 \sin \sqrt{x} + c$.

(A) Let $I = \int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx$.
Substitute $t = a \cos^2 x + b \sin^2 x$.
Then $dt = (a \cdot 2 \cos x(-\sin x) + b \cdot 2 \sin x \cos x) dx = 2(b - a) \sin x \cos x dx$.
So,$\sin x \cos x dx = \frac{dt}{2(b - a)}$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t} \cdot \frac{dt}{2(b - a)} = \frac{1}{2(b - a)} \int \frac{1}{t} dt$.
$I = \frac{1}{2(b - a)} \log |t| + C$.
Substituting back $t = a \cos^2 x + b \sin^2 x$,we get:
$I = \frac{1}{2(b - a)} \log |a \cos^2 x + b \sin^2 x| + C$.

(C) Let $t = \tan^{-1} x$.
Then,differentiating with respect to $x$,we get $dt = \frac{1}{1 + x^2} dx$.
Substituting these into the integral,we have:
$\int \frac{e^{\tan^{-1} x}}{1 + x^2} dx = \int e^t dt$.
The integral of $e^t$ is $e^t + c$.
Substituting back $t = \tan^{-1} x$,we get $e^{\tan^{-1} x} + c$.

(B) To solve the integral $\int \frac{1}{x(\log x)^2} \, dx$,we use the method of substitution.
Let $t = \log x$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $dt = \frac{1}{x} \, dx$.
Substituting these into the integral,we get:
$\int \frac{1}{t^2} \, dt = \int t^{-2} \, dt$.
Using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$,we have:
$\int t^{-2} \, dt = \frac{t^{-2+1}}{-2+1} + c = \frac{t^{-1}}{-1} + c = -\frac{1}{t} + c$.
Finally,substituting $t = \log x$ back into the expression,we get:
$-\frac{1}{\log x} + c$.

(C) Let $I = \int \frac{1}{\sqrt{x}} \tan^4(\sqrt{x}) \sec^2(\sqrt{x}) \, dx$.
Substitute $t = \tan(\sqrt{x})$.
Then,$dt = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \, dx$,which implies $\frac{\sec^2(\sqrt{x})}{\sqrt{x}} \, dx = 2 \, dt$.
Substituting these into the integral,we get:
$I = \int t^4 \cdot 2 \, dt = 2 \int t^4 \, dt$.
Integrating $t^4$ with respect to $t$,we get $2 \cdot \frac{t^5}{5} + c$.
Substituting back $t = \tan(\sqrt{x})$,we obtain $I = \frac{2}{5} \tan^5(\sqrt{x}) + c$.

(A) Let $a^x = t$. Then,differentiating both sides with respect to $x$,we get $a^x \log_e a \, dx = dt$,which implies $a^x \, dx = \frac{dt}{\log_e a}$.
Substituting these into the integral:
$\int \frac{a^x}{\sqrt{1 - a^{2x}}} dx = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{dt}{\log_e a}$
$= \frac{1}{\log_e a} \int \frac{dt}{\sqrt{1 - t^2}}$
$= \frac{1}{\log_e a} \sin^{-1}(t) + c$
$= \frac{\sin^{-1}(a^x)}{\log_e a} + c$.

(B) Let $I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx$.
Multiply the numerator and denominator by $\cos x$:
$I = \int \frac{\sqrt{\tan x}}{\sin x \cos x \cdot \frac{\cos x}{\cos x}} \, dx = \int \frac{\sqrt{\tan x}}{\tan x \cos^2 x} \, dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$.
Substituting these into the integral:
$I = \int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} + c = 2\sqrt{t} + c$.
Replacing $t$ with $\tan x$,we get $I = 2\sqrt{\tan x} + c$.

(A) Let $I = \int \frac{\sin 2x}{a^2 + b^2 \sin^2 x} \, dx$.
Substitute $t = a^2 + b^2 \sin^2 x$.
Differentiating with respect to $x$,we get $dt = b^2 (2 \sin x \cos x) \, dx = b^2 \sin 2x \, dx$.
Therefore,$\sin 2x \, dx = \frac{dt}{b^2}$.
Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{dt}{b^2} = \frac{1}{b^2} \int \frac{1}{t} \, dt$.
Integrating,we get $I = \frac{1}{b^2} \log |t| + c$.
Substituting $t$ back,we get $I = \frac{1}{b^2} \log(a^2 + b^2 \sin^2 x) + c$.

(C) Let $t = 1 + \log x$.
Then,the derivative is $dt = \frac{1}{x} \, dx$.
Substituting these into the integral:
$\int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt$.
Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$:
$= \frac{t^{1/2}}{1/2} + c = 2t^{1/2} + c$.
Substituting $t = 1 + \log x$ back,we get:
$= 2\sqrt{1 + \log x} + c$.

(C) Let $I = \int \frac{\sec^2 x}{1 + \tan x} \, dx$.
Substitute $t = 1 + \tan x$.
Then,the derivative is $dt = \sec^2 x \, dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t} \, dt$.
Integrating with respect to $t$,we obtain:
$I = \log |t| + c$.
Replacing $t$ with $1 + \tan x$,we get:
$I = \log |1 + \tan x| + c$.

(B) Let $I = \int {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \,dx$.
Divide the numerator and denominator by ${e^x}$:
$I = \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \,dx$.
Let $t = {e^x} + {e^{ - x}}$.
Then $dt = ({e^x} - {e^{ - x}}) \,dx$.
Substituting these into the integral:
$I = \int {\frac{1}{t}} \,dt = \log |t| + c$.
Substituting back $t = {e^x} + {e^{ - x}}$:
$I = \log ({e^x} + {e^{ - x}}) + c$.
Since ${e^x} + {e^{ - x}} = \frac{{{e^{2x}} + 1}}{{{e^x}}}$,we have:
$I = \log \left( \frac{{{e^{2x}} + 1}}{{{e^x}}} \right) + c = \log ({e^{2x}} + 1) - \log ({e^x}) + c = \log ({e^{2x}} + 1) - x + c$.

(A) Let $I = \int \frac{\csc x}{\log \tan \frac{x}{2}} \, dx$.
Substitute $t = \log \tan \frac{x}{2}$.
Then,differentiating with respect to $x$,we get:
$\frac{dt}{dx} = \frac{1}{\tan \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \csc x$.
Thus,$\csc x \, dx = dt$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t} \, dt = \log |t| + c$.
Replacing $t$ with $\log \tan \frac{x}{2}$,we obtain:
$I = \log \left| \log \tan \frac{x}{2} \right| + c$.

(B) We have the integral $I = \int \frac{1}{\cos^2 x (1 - \tan x)^2} dx$.
Since $\frac{1}{\cos^2 x} = \sec^2 x$,the integral becomes $I = \int \frac{\sec^2 x}{(1 - \tan x)^2} dx$.
Let $u = 1 - \tan x$. Then $du = -\sec^2 x dx$,which implies $\sec^2 x dx = -du$.
Substituting these into the integral,we get $I = \int \frac{-du}{u^2} = -\int u^{-2} du$.
Integrating,we get $I = -(\frac{u^{-1}}{-1}) + c = \frac{1}{u} + c$.
Substituting back $u = 1 - \tan x$,we get $I = \frac{1}{1 - \tan x} + c$.

(B) Let $I = \int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx$.
Substitute $t = {x^{10}} + {10^x}$.
Differentiating with respect to $x$,we get $dt = (10{x^9} + {10^x}{\log _e}10) \;dx$.
Substituting these into the integral,we get $I = \int {\frac{1}{t}} \;dt$.
Integrating,we get $I = \log |t| + c$.
Substituting back the value of $t$,we get $I = \log |{x^{10}} + {10^x}| + c$.

(A) We have the integral $I = \int \frac{1}{(e^x + e^{-x})^2} \, dx$.
Multiply the numerator and denominator by $e^{2x}$:
$I = \int \frac{e^{2x}}{(e^x \cdot e^x + e^x \cdot e^{-x})^2} \, dx = \int \frac{e^{2x}}{(e^{2x} + 1)^2} \, dx$.
Let $t = e^{2x} + 1$. Then $dt = 2e^{2x} \, dx$,which implies $e^{2x} \, dx = \frac{1}{2} \, dt$.
Substituting these into the integral:
$I = \int \frac{1}{t^2} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int t^{-2} \, dt$.
Integrating,we get $I = \frac{1}{2} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{2t} + c$.
Substituting back $t = e^{2x} + 1$,we get $I = -\frac{1}{2(e^{2x} + 1)} + c$.

(C) We know that $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.
Substituting this into the integral:
$\int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \, dx = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$.
Let $t = \cos x + \sin x$. Then $dt = (-\sin x + \cos x) \, dx = (\cos x - \sin x) \, dx$.
The integral becomes $\int \frac{1}{t} \, dt = \log |t| + c$.
Substituting back $t = \cos x + \sin x$,we get $\log |\cos x + \sin x| + c$.

(C) Let $I = \int \frac{\tan(\log x)}{x} \, dx$.
Substitute $\log x = t$.
Then,differentiating with respect to $x$,we get $\frac{1}{x} \, dx = dt$.
Substituting these into the integral,we get:
$I = \int \tan(t) \, dt$.
The integral of $\tan(t)$ is $\log |\sec(t)| + c$.
Substituting $t = \log x$ back,we get:
$I = \log |\sec(\log x)| + c$.

(B) Let $t = x^2$.
Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$.
Substituting these into the integral:
$\int \frac{x}{1 + x^4} \, dx = \int \frac{1}{1 + (x^2)^2} \cdot (x \, dx) = \int \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt$.
$= \frac{1}{2} \int \frac{1}{1 + t^2} \, dt$.
Using the standard integral formula $\int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) + c$,we get:
$= \frac{1}{2} \tan^{-1}(t) + c$.
Substituting $t = x^2$ back,the final result is $\frac{1}{2} \tan^{-1}(x^2) + c$.

(A) Let $I = \int \frac{1}{\sqrt{1 - e^{2x}}} \, dx$.
Multiply the numerator and denominator by $e^{-x}$:
$I = \int \frac{e^{-x}}{\sqrt{e^{-2x} - 1}} \, dx$.
Substitute $t = e^{-x}$,then $dt = -e^{-x} \, dx$,which implies $e^{-x} \, dx = -dt$.
$I = -\int \frac{1}{\sqrt{t^2 - 1}} \, dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{t^2 - a^2}} \, dt = \log|t + \sqrt{t^2 - a^2}| + c$:
$I = -\log|t + \sqrt{t^2 - 1}| + c$.
Substitute $t = e^{-x}$ back:
$I = -\log|e^{-x} + \sqrt{e^{-2x} - 1}| + c$.
$I = -\log|\frac{1}{e^x} + \frac{\sqrt{1 - e^{2x}}}{e^x}| + c$.
$I = -\log|\frac{1 + \sqrt{1 - e^{2x}}}{e^x}| + c$.
$I = -\log(1 + \sqrt{1 - e^{2x}}) + \log(e^x) + c$.
$I = x - \log(1 + \sqrt{1 - e^{2x}}) + c$.

(A) Let $I = \int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx$.
We can rewrite the integral as:
$I = \int \frac{3x^2}{\sqrt{3^2 - (4x^3)^2}} \, dx$.
Let $t = 4x^3$. Then $dt = 12x^2 \, dx$,which implies $3x^2 \, dx = \frac{1}{4} dt$.
Substituting these into the integral:
$I = \frac{1}{4} \int \frac{dt}{\sqrt{3^2 - t^2}}$.
Using the standard formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + c$:
$I = \frac{1}{4} \sin^{-1} \left( \frac{t}{3} \right) + c$.
Substituting $t = 4x^3$ back:
$I = \frac{1}{4} \sin^{-1} \left( \frac{4x^3}{3} \right) + c$.

(B) Let $I = \int \cos x \sqrt{4 - \sin^2 x} \; dx$.
Substitute $\sin x = t$,then $\cos x \; dx = dt$.
The integral becomes $I = \int \sqrt{4 - t^2} \; dt = \int \sqrt{2^2 - t^2} \; dt$.
Using the standard formula $\int \sqrt{a^2 - x^2} \; dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + c$,we get:
$I = \frac{t}{2} \sqrt{4 - t^2} + \frac{4}{2} \sin^{-1} \left( \frac{t}{2} \right) + c$.
Substituting $t = \sin x$ back into the expression:
$I = \frac{1}{2} \sin x \sqrt{4 - \sin^2 x} + 2 \sin^{-1} \left( \frac{\sin x}{2} \right) + c$.

(D) Let $I = \int x^2 (3)^{x^3 + 1} dx$.
We know that $3^{x^3 + 1} = 3 \cdot 3^{x^3}$.
So,$I = \int x^2 \cdot 3 \cdot 3^{x^3} dx = 3 \int x^2 \cdot 3^{x^3} dx$.
Let $t = x^3$. Then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{dt}{3}$.
Substituting these into the integral:
$I = 3 \int 3^t \cdot \frac{dt}{3} = \int 3^t dt$.
Using the formula $\int a^t dt = \frac{a^t}{\ln a} + c$,we get:
$I = \frac{3^t}{\ln 3} + c$.
Substituting $t = x^3$ back,we get:
$I = \frac{3^{x^3}}{\ln 3} + c$.
Since the original expression was $3^{x^3+1}$,note that $\frac{3^{x^3+1}}{3 \ln 3} = \frac{3 \cdot 3^{x^3}}{3 \ln 3} = \frac{3^{x^3}}{\ln 3}$.
Thus,the correct form is $\frac{3^{x^3}}{\ln 3} + c$.

(B) We have $I = \int \sec^{2/3} x \csc^{4/3} x \, dx = \int \frac{1}{\cos^{2/3} x \sin^{4/3} x} \, dx$.
Multiplying the numerator and denominator by $\sec^2 x$,we get:
$I = \int \frac{\sec^2 x}{\frac{\cos^{2/3} x \sin^{4/3} x}{\cos^2 x}} \, dx = \int \frac{\sec^2 x}{\frac{\sin^{4/3} x}{\cos^{4/3} x}} \, dx = \int \frac{\sec^2 x}{(\tan x)^{4/3}} \, dx$.
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
Substituting these into the integral,we get:
$I = \int t^{-4/3} \, dt = \frac{t^{-4/3 + 1}}{-4/3 + 1} + c = \frac{t^{-1/3}}{-1/3} + c = -3t^{-1/3} + c$.
Replacing $t$ with $\tan x$,we get $I = -3(\tan x)^{-1/3} + c$.

(A) We have $I = \int \cos^5 x \, dx$.
We can write this as $I = \int \cos^4 x \cdot \cos x \, dx = \int (1 - \sin^2 x)^2 \cos x \, dx$.
Let $t = \sin x$,then $dt = \cos x \, dx$.
Substituting these into the integral,we get $I = \int (1 - t^2)^2 \, dt$.
Expanding the integrand,$I = \int (1 - 2t^2 + t^4) \, dt$.
Integrating term by term,$I = t - \frac{2t^3}{3} + \frac{t^5}{5} + c$.
Substituting $t = \sin x$ back,we get $I = \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + c$.

(C) Let $I = \int \frac{d\theta}{\sin \theta \cos^3 \theta}$.
Divide the numerator and denominator by $\cos^4 \theta$:
$I = \int \frac{\sec^4 \theta \, d\theta}{\tan \theta} = \int \frac{\sec^2 \theta (1 + \tan^2 \theta) \, d\theta}{\tan \theta}$.
Substitute $t = \tan \theta$,so $dt = \sec^2 \theta \, d\theta$:
$I = \int \frac{1 + t^2}{t} \, dt = \int \left( \frac{1}{t} + t \right) \, dt$.
Integrating with respect to $t$:
$I = \log |t| + \frac{t^2}{2} + c$.
Substituting back $t = \tan \theta$:
$I = \log |\tan \theta| + \frac{\tan^2 \theta}{2} + c$.