Integration by substitution Questions in English

(B) Let $t = \cos^{-1} x$.
Then,differentiating both sides with respect to $x$,we get $dt = -\frac{1}{\sqrt{1 - x^2}} dx$,which implies $\frac{dx}{\sqrt{1 - x^2}} = -dt$.
Substituting these into the integral:
$\int \frac{1}{(\cos^{-1} x) \sqrt{1 - x^2}} dx = \int \frac{1}{t} (-dt) = -\int \frac{1}{t} dt$.
Integrating,we get $-\log|t| + c$.
Substituting $t = \cos^{-1} x$ back,the result is $-\log(\cos^{-1} x) + c$.

(B) The simplest way is to substitute $3x^2 + 5 = t$.
Let $t = 3x^2 + 5$. Then $dt = 6x dx$,which implies $x dx = \frac{dt}{6}$.
Also,$x^2 = \frac{t - 5}{3}$.
Substituting these into the integral:
$\int x^3 e^{3x^2 + 5} dx = \int x^2 \cdot e^{3x^2 + 5} \cdot x dx = \int \left( \frac{t - 5}{3} \right) e^t \cdot \frac{dt}{6} = \frac{1}{18} \int (t - 5) e^t dt$.
$= \frac{1}{18} \left[ \int t e^t dt - 5 \int e^t dt \right] = \frac{1}{18} [ (t e^t - e^t) - 5 e^t ] + c = \frac{1}{18} (t e^t - 6 e^t) + c$.
Substituting back $t = 3x^2 + 5$:
$= \frac{1}{18} (3x^2 + 5) e^{3x^2 + 5} - \frac{6}{18} e^{3x^2 + 5} + c = \frac{1}{18} (3x^2 + 5) e^{3x^2 + 5} - \frac{1}{3} e^{3x^2 + 5} + c$.

(C) Let $I = \int \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \, dx$.
To simplify the integral,we use the substitution $\tan x = t$.
Differentiating both sides with respect to $x$,we get $\sec^2 x \, dx = dt$.
Substituting these into the integral,we get $I = \int \frac{dt}{(1 + t)(2 + t)}$.
This is a standard form that can be solved using partial fractions.
Thus,the most suitable substitution is $\tan x = t$.

(B) Let $I = \int \frac{\csc^2 x}{1 + \cot x} dx$.
Substitute $1 + \cot x = t$.
Differentiating both sides with respect to $x$,we get $-\csc^2 x dx = dt$,which implies $\csc^2 x dx = -dt$.
Substituting these into the integral,we get $I = \int \frac{-dt}{t}$.
Integrating,we get $I = -\log|t| + c$.
Replacing $t$ with $1 + \cot x$,we get $I = -\log|1 + \cot x| + c$.

(B) To solve the integral $\int \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx$,we use the method of substitution.
Let $t = \sqrt{x}$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{2\sqrt{x}}$,which implies $\frac{dx}{\sqrt{x}} = 2 \, dt$.
Substituting these into the integral,we get:
$\int \sin(t) \cdot 2 \, dt = 2 \int \sin(t) \, dt$.
The integral of $\sin(t)$ is $-\cos(t)$.
Therefore,$2(-\cos(t)) + c = -2 \cos(t) + c$.
Substituting back $t = \sqrt{x}$,we get $-2 \cos \sqrt{x} + c$.

(D) Let $I = \int e^x \tan^2(e^x) \, dx$.
Substitute $t = e^x$,then $dt = e^x \, dx$.
Substituting these into the integral,we get:
$I = \int \tan^2(t) \, dt$.
Using the trigonometric identity $\tan^2(t) = \sec^2(t) - 1$,we have:
$I = \int (\sec^2(t) - 1) \, dt$.
Integrating term by term:
$I = \tan(t) - t + c$.
Substituting $t = e^x$ back into the expression:
$I = \tan(e^x) - e^x + c$.

(A) Given integral: $I = \int \frac{dx}{e^{-2x}(e^{2x} + 1)^2}$
Since $\frac{1}{e^{-2x}} = e^{2x}$,the integral becomes:
$I = \int \frac{e^{2x} dx}{(e^{2x} + 1)^2}$
Let $t = e^{2x} + 1$. Then,differentiating with respect to $x$,we get $dt = 2e^{2x} dx$,which implies $e^{2x} dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{1}{t^2} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{-2} dt$
Using the power rule for integration $\int t^n dt = \frac{t^{n+1}}{n+1}$:
$I = \frac{1}{2} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{2t} + c$
Substituting back $t = e^{2x} + 1$:
$I = \frac{-1}{2(e^{2x} + 1)} + c$

(C) Let $I = \int \frac{dx}{x\sqrt{1 - (\log x)^2}}$.
Substitute $t = \log x$.
Then,the derivative is $dt = \frac{1}{x} dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{\sqrt{1 - t^2}}$.
Using the standard integral formula $\int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$,we obtain:
$I = \sin^{-1}(t) + c$.
Substituting back $t = \log x$,we get:
$I = \sin^{-1}(\log x) + c$.

(C) The substitution ${x^2} = t$ implies $2x \,dx = dt$,or $x \,dx = \frac{1}{2} \,dt$.
For option $(c)$,we have $\int {x^3 \cos(x^2)} \,dx = \int {x^2 \cdot \cos(x^2) \cdot x \,dx}$.
Substituting ${x^2} = t$ and $x \,dx = \frac{1}{2} \,dt$,we get $\frac{1}{2} \int {t \cos t \,dt}$.
Using integration by parts: $\frac{1}{2} [t \sin t - \int \sin t \,dt] = \frac{1}{2} (t \sin t + \cos t) + C$.
Substituting back $t = x^2$,we get $\frac{1}{2} (x^2 \sin(x^2) + \cos(x^2)) + C$.
Thus,option $(c)$ is correct.

(A) Let $I = \int \tan x \sec^2 x \sqrt{1 - \tan^2 x} \; dx$.
Substitute $\tan x = t$,then $\sec^2 x \; dx = dt$.
The integral becomes $\int t \sqrt{1 - t^2} \; dt$.
Now,substitute $1 - t^2 = u$,then $-2t \; dt = du$,or $t \; dt = -\frac{1}{2} du$.
Substituting these into the integral: $I = \int \sqrt{u} \left(-\frac{1}{2}\right) du = -\frac{1}{2} \int u^{1/2} \; du$.
Integrating with respect to $u$: $I = -\frac{1}{2} \left(\frac{u^{3/2}}{3/2}\right) + c = -\frac{1}{3} u^{3/2} + c$.
Substituting $u = 1 - t^2 = 1 - \tan^2 x$ back: $I = -\frac{1}{3}(1 - \tan^2 x)^{3/2} + c$.

(B) Let $e^x = t$. Then,differentiating both sides with respect to $x$,we get $e^x \, dx = dt$.
Substituting these into the integral,we have:
$\int \frac{dt}{\sqrt{1 - t^2}}$.
We know that $\int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$ or $-\cos^{-1}(t) + c$.
Since the options provided use the cosine inverse function,we use the identity $\int \frac{dt}{\sqrt{1 - t^2}} = -\cos^{-1}(t) + c$.
Substituting $t = e^x$ back,we get $-\cos^{-1}(e^x) + c$.

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

(B) Let $I = \int \frac{\sin x}{(a + b \cos x)^2} \, dx$.
Substitute $t = a + b \cos x$.
Then,$dt = -b \sin x \, dx$,which implies $\sin x \, dx = -\frac{dt}{b}$.
Substituting these into the integral:
$I = \int \frac{1}{t^2} \left(-\frac{dt}{b}\right) = -\frac{1}{b} \int t^{-2} \, dt$.
Integrating $t^{-2}$ gives $-t^{-1} = -\frac{1}{t}$.
Thus,$I = -\frac{1}{b} \left(-\frac{1}{t}\right) + c = \frac{1}{bt} + c$.
Substituting back $t = a + b \cos x$,we get $I = \frac{1}{b(a + b \cos x)} + c$.

(B) Given integral: $I = \int \frac{1}{x^3} [\log x^x]^2 \, dx$
Since $\log x^x = x \log x$,we substitute this into the integral:
$I = \int \frac{1}{x^3} (x \log x)^2 \, dx$
$I = \int \frac{1}{x^3} (x^2 (\log x)^2) \, dx$
$I = \int \frac{1}{x} (\log x)^2 \, dx$
Now,let $t = \log x$,then $dt = \frac{1}{x} \, dx$.
Substituting these into the integral:
$I = \int t^2 \, dt = \frac{t^3}{3} + c$
Replacing $t$ with $\log x$:
$I = \frac{1}{3}(\log x)^3 + c$

(A) 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:
$\int \sec^2(\log x) \cdot \frac{1}{x} \, dx = \int \sec^2(t) \, dt$.
The integral of $\sec^2(t)$ is $\tan(t) + c$.
Replacing $t$ with $\log x$,we get $\tan(\log x) + c$.

(B) Let $I = \int \frac{dx}{x \log x \log(\log x)}$.
Substitute $\log x = t$. Then,$\frac{1}{x} dx = dt$.
The integral becomes $\int \frac{dt}{t \log t}$.
Now,substitute $\log t = z$. Then,$\frac{1}{t} dt = dz$.
The integral becomes $\int \frac{dz}{z} = \log|z| + c$.
Substituting back the values,we get $\log|\log t| + c = \log|\log(\log x)| + c$.

(A) Let $t = \tan x$. Then,$dt = \sec^2 x \, dx$.
Substituting these into the integral,we get:
$\int \frac{dt}{\sqrt{t^2 + 2^2}}$.
Using the standard integration formula $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log |x + \sqrt{x^2 + a^2}| + c$,we obtain:
$\log |t + \sqrt{t^2 + 2^2}| + c$.
Substituting $t = \tan x$ back,the final result is:
$\log |\tan x + \sqrt{\tan^2 x + 4}| + c$.

(B) Let $t = \tan^{-1}(x^2)$.
Then,differentiating both sides with respect to $x$,we get $dt = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) \, dx = \frac{2x}{1 + x^4} \, dx$.
Substituting these into the integral,we have:
$\int \tan^{-1}(x^2) \cdot \frac{2x}{1 + x^4} \, dx = \int t \, dt$.
Integrating with respect to $t$,we get $\frac{t^2}{2} + c$.
Substituting back the value of $t$,we obtain $\frac{1}{2} (\tan^{-1}(x^2))^2 + c$.

(C) Let $I = \int \frac{x^3}{\sqrt{1 - x^8}} \, dx$.
Rewrite the integral as $I = \int \frac{x^3}{\sqrt{1 - (x^4)^2}} \, dx$.
Substitute $t = x^4$,then $dt = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{1}{4} \, dt$.
Substituting these into the integral,we get $I = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{1}{4} \, dt$.
$I = \frac{1}{4} \int \frac{1}{\sqrt{1 - t^2}} \, dt$.
Using the standard integral $\int \frac{1}{\sqrt{1 - t^2}} \, dt = \sin^{-1}(t) + c$,we get $I = \frac{1}{4} \sin^{-1}(t) + c$.
Substituting back $t = x^4$,we obtain $I = \frac{1}{4} \sin^{-1}(x^4) + c$.

(A) Let $t = \cos(x^2)$.
Then,differentiating with respect to $x$,we get $dt = -\sin(x^2) \cdot 2x \, dx$,which implies $2x \sin(x^2) \, dx = -dt$.
Substituting these into the integral:
$\int 2x \cos^3(x^2) \sin(x^2) \, dx = \int t^3 (-dt)$
$= -\int t^3 \, dt$
$= -\frac{t^4}{4} + c$
Substituting $t = \cos(x^2)$ back into the expression:
$= -\frac{1}{4} \cos^4(x^2) + c$.

(A) To solve the integral $\int \sec^4 x \tan x \; dx$,we can rewrite the integrand as:
$\int \sec^3 x (\sec x \tan x) \; dx$
Let $t = \sec x$.
Then,the derivative is $dt = \sec x \tan x \; dx$.
Substituting these into the integral,we get:
$\int t^3 \; dt$
Integrating with respect to $t$,we obtain:
$\frac{t^4}{4} + c$
Finally,substituting $t = \sec x$ back into the expression,we get:
$\frac{1}{4} \sec^4 x + c$

(A) Let $I = \int {{e^{ - x}}{{\csc }^2}(2{e^{ - x}} + 5)} \,dx$.
Substitute $t = 2{e^{ - x}} + 5$.
Then,$dt = -2{e^{ - x}}dx$,which implies ${e^{ - x}}dx = -\frac{1}{2}dt$.
Substituting these into the integral:
$I = \int {\csc^2(t) \cdot (-\frac{1}{2}dt)} = -\frac{1}{2} \int \csc^2(t) dt$.
Since $\int \csc^2(t) dt = -\cot(t) + c$,we get:
$I = -\frac{1}{2} (-\cot(t)) + c = \frac{1}{2} \cot(t) + c$.
Substituting back $t = 2{e^{ - x}} + 5$,we obtain:
$I = \frac{1}{2} \cot(2{e^{ - x}} + 5) + c$.

(B) To solve the integral $\int \sin^3 x \cdot \cos x \, dx$,we use the method of substitution.
Let $t = \sin x$.
Then,differentiating both sides with respect to $x$,we get $dt = \cos x \, dx$.
Substituting these into the integral,we have $\int t^3 \, dt$.
The integral of $t^3$ with respect to $t$ is $\frac{t^4}{4} + c$.
Substituting back $t = \sin x$,we get the final result $\frac{\sin^4 x}{4} + c$.

(C) Let $I = \int \frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} \, dx$.
Consider the denominator $t = x^2 + \sin 2x + 2x$.
Differentiating with respect to $x$,we get $\frac{dt}{dx} = 2x + 2\cos 2x + 2 = 2(x + \cos 2x + 1)$.
Thus,$dt = 2(x + \cos 2x + 1) \, dx$,which implies $(x + \cos 2x + 1) \, dx = \frac{1}{2} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{t} \, dt$.
Integrating,we get $I = \frac{1}{2} \log |t| + c$.
Substituting back $t$,we get $I = \frac{1}{2} \log |x^2 + \sin 2x + 2x| + c$.

(A) Let $t = x + \log \sec x$.
Then,the derivative of $t$ with respect to $x$ is:
$\frac{dt}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\log \sec x) = 1 + \frac{1}{\sec x} \cdot (\sec x \tan x) = 1 + \tan x$.
So,$dt = (1 + \tan x) \, dx$.
Substituting these into the integral:
$\int \frac{1 + \tan x}{x + \log \sec x} \, dx = \int \frac{1}{t} \, dt = \log |t| + c$.
Replacing $t$ with the original expression:
$\log |x + \log \sec x| + c$.

(C) Let $I = \int \frac{(x + 1)(x + \log x)^2}{x} \, dx$.
We can rewrite the integrand as $\int (x + \log x)^2 \left( \frac{x + 1}{x} \right) \, dx = \int (x + \log x)^2 \left( 1 + \frac{1}{x} \right) \, dx$.
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 get $I = \int t^2 \, dt$.
Integrating $t^2$ with respect to $t$,we get $I = \frac{t^3}{3} + c$.
Substituting back $t = x + \log x$,we get $I = \frac{1}{3}(x + \log x)^3 + c$.

(A) We have the integral $I = \int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx$.
Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$,the integral becomes:
$I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) \, dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t^2} \, dt = \int t^{-2} \, dt = \frac{t^{-1}}{-1} + c = -\frac{1}{t} + c$.
Substituting back $t = \sin x + \cos x$,we get:
$I = -\frac{1}{\sin x + \cos x} + c$.

(C) Let $I = \int {{x^3}\sqrt {3 + 5{x^4}} } \;dx$.
Substitute $3 + 5{x^4} = t$.
Differentiating both sides with respect to $x$,we get $20{x^3}dx = dt$,which implies ${x^3}dx = \frac{1}{20}dt$.
Substituting these into the integral:
$I = \int \sqrt{t} \cdot \frac{1}{20} dt = \frac{1}{20} \int t^{1/2} dt$.
Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + C$:
$I = \frac{1}{20} \cdot \frac{t^{3/2}}{3/2} + C = \frac{1}{20} \cdot \frac{2}{3} t^{3/2} + C = \frac{1}{30} t^{3/2} + C$.
Substituting back $t = 3 + 5{x^4}$,we get:
$I = \frac{1}{30}{(3 + 5{x^4})^{3/2}} + C$.

(B) Let $I = \int \sqrt{\frac{x}{a^3 - x^3}} \, dx$.
Substitute $x = a(\sin \theta)^{2/3}$.
Then $dx = a \cdot \frac{2}{3}(\sin \theta)^{-1/3} \cdot \cos \theta \, d\theta$.
Substituting these into the integral:
$I = \int \sqrt{\frac{a(\sin \theta)^{2/3}}{a^3 - a^3(\sin \theta)^2}} \cdot \frac{2}{3}a(\sin \theta)^{-1/3} \cos \theta \, d\theta$
$I = \int \sqrt{\frac{a(\sin \theta)^{2/3}}{a^3(1 - \sin^2 \theta)}} \cdot \frac{2}{3}a(\sin \theta)^{-1/3} \cos \theta \, d\theta$
$I = \int \frac{a^{1/2}(\sin \theta)^{1/3}}{a^{3/2} \cos \theta} \cdot \frac{2}{3}a(\sin \theta)^{-1/3} \cos \theta \, d\theta$
$I = \frac{2}{3} \int d\theta = \frac{2}{3} \theta + c$.
Since $x = a(\sin \theta)^{2/3}$,we have $(\frac{x}{a})^{3/2} = \sin \theta$,so $\theta = \sin^{-1}(\frac{x}{a})^{3/2}$.
Thus,$I = \frac{2}{3} \sin^{-1}(\frac{x}{a})^{3/2} + c$.

(A) Let $I = \int \frac{1}{x \cos^2(1 + \log x)} \, dx$.
Substitute $t = 1 + \log x$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $dt = \frac{1}{x} \, dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{\cos^2 t} \, dt = \int \sec^2 t \, dt$.
Since the integral of $\sec^2 t$ is $\tan t + c$,we have:
$I = \tan t + c$.
Substituting back $t = 1 + \log x$,we get:
$I = \tan(1 + \log x) + c$.

(A) Let $x = \tan \theta$. Then $dx = \sec^2 \theta \, d\theta$.
Substituting these into the integral:
$\int \frac{1}{x^2 \sqrt{1 + x^2}} \, dx = \int \frac{\sec^2 \theta \, d\theta}{\tan^2 \theta \sqrt{1 + \tan^2 \theta}} = \int \frac{\sec^2 \theta \, d\theta}{\tan^2 \theta \sec \theta} = \int \frac{\sec \theta}{\tan^2 \theta} \, d\theta$.
Simplifying the expression:
$\int \frac{1/\cos \theta}{\sin^2 \theta / \cos^2 \theta} \, d\theta = \int \frac{\cos \theta}{\sin^2 \theta} \, d\theta = \int \csc \theta \cot \theta \, d\theta$.
The integral of $\csc \theta \cot \theta$ is $-\csc \theta + c$.
Since $x = \tan \theta$,we have $\tan \theta = \frac{x}{1}$,so $\csc \theta = \frac{\sqrt{1 + x^2}}{x}$.
Thus,the final result is $-\frac{\sqrt{1 + x^2}}{x} + c$.

(A) Let $I = \int \frac{\log (x + \sqrt {1 + x^2})}{\sqrt {1 + x^2}} \, dx$.
Substitute $t = \log (x + \sqrt {1 + x^2})$.
Differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = \frac{1}{x + \sqrt {1 + x^2}} \cdot (1 + \frac{1}{2\sqrt {1 + x^2}} \cdot 2x) = \frac{1}{x + \sqrt {1 + x^2}} \cdot (1 + \frac{x}{\sqrt {1 + x^2}}) = \frac{1}{x + \sqrt {1 + x^2}} \cdot \frac{\sqrt {1 + x^2} + x}{\sqrt {1 + x^2}} = \frac{1}{\sqrt {1 + x^2}}$.
Thus,$dt = \frac{dx}{\sqrt {1 + x^2}}$.
Substituting these into the integral:
$I = \int t \, dt = \frac{t^2}{2} + c$.
Replacing $t$ with its original value:
$I = \frac{1}{2}[\log (x + \sqrt {1 + x^2})]^2 + c$.

(A) To evaluate the integral $I = \int {{e^x}\sin ({e^x})} \,dx$,we use the method of substitution.
Let $t = {e^x}$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = {e^x}$,which implies $dt = {e^x}dx$.
Substituting these into the integral,we get:
$I = \int \sin(t) \,dt$.
The integral of $\sin(t)$ is $-\cos(t) + c$.
Substituting back $t = {e^x}$,we get $I = -\cos({e^x}) + c$.

(D) Let $I = \int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} $.
We can write $x^5 dx$ as $x^3 \cdot x^2 dx$.
Let $1 + x^3 = t^2$. Then $3x^2 dx = 2t dt$,which implies $x^2 dx = \frac{2}{3}t dt$.
Also,$x^3 = t^2 - 1$.
Substituting these into the integral:
$I = \int {\frac{{(t^2 - 1)}}{t} \cdot \frac{2}{3}t dt} = \frac{2}{3} \int {(t^2 - 1) dt} $.
Integrating with respect to $t$:
$I = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + C = \frac{2}{9} t(t^2 - 3) + C$.
Since $t = \sqrt{1 + x^3}$ and $t^2 = 1 + x^3$,we have:
$I = \frac{2}{9} \sqrt{1 + x^3} (1 + x^3 - 3) + C = \frac{2}{9} \sqrt{1 + x^3} (x^3 - 2) + C$.

(A) We have $I = \int {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\,dx}$.
Taking $x^4$ common from the bracket,we get $I = \int {\frac{{{{[x^4(1 - 1/x^3)]}^{1/4}}}}{{{x^5}}}\,dx} = \int {\frac{{x{{(1 - 1/x^3)}^{1/4}}}}{{{x^5}}}\,dx} = \int {\frac{{{{(1 - 1/x^3)}^{1/4}}}}{{{x^4}}}\,dx}$.
Let $1 - \frac{1}{{{x^3}}} = t$. Then,differentiating both sides with respect to $x$,we get $\frac{3}{{{x^4}}}dx = dt$,which implies $\frac{1}{{{x^4}}}dx = \frac{1}{3}dt$.
Substituting these into the integral,we get $I = \int {{t^{1/4}} \cdot \frac{1}{3}dt} = \frac{1}{3} \cdot \frac{{{t^{5/4}}}}{{5/4}} + c = \frac{1}{3} \cdot \frac{4}{5} {t^{5/4}} + c = \frac{4}{{15}}{t^{5/4}} + c$.
Substituting back $t = 1 - \frac{1}{{{x^3}}}$,we get $I = \frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$.

(A) Let $I = \int e^x \sec^2(e^x) \, dx$.
Substitute $t = e^x$.
Then,the derivative is $dt = e^x \, dx$.
Substituting these into the integral,we get:
$I = \int \sec^2(t) \, dt$.
Since the integral of $\sec^2(t)$ is $\tan(t)$,we have:
$I = \tan(t) + k$.
Substituting $t = e^x$ back into the expression,we get:
$I = \tan(e^x) + k$.

(D) Let $I = \int t e^{-3t^2} \, dt$.
Substitute $z = -3t^2$,then $dz = -6t \, dt$,which implies $t \, dt = -\frac{1}{6} \, dz$.
Substituting these into the integral:
$I = \int e^z \left(-\frac{1}{6}\right) \, dz = -\frac{1}{6} \int e^z \, dz$.
Integrating $e^z$ gives $e^z$:
$I = -\frac{1}{6} e^z + c$.
Substituting back $z = -3t^2$:
$I = -\frac{1}{6} e^{-3t^2} + c$.

(B) Let $I = \int \frac{1}{(1 + x)\sqrt{x}} \, dx$.
We can rewrite the integral as $I = \int \frac{1}{(1 + (\sqrt{x})^2)\sqrt{x}} \, dx$.
Let $\sqrt{x} = t$. Then,differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} \, dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 \, dt$.
Substituting these into the integral,we get $I = \int \frac{2 \, dt}{1 + t^2}$.
Integrating,we obtain $I = 2\tan^{-1}(t) + A$.
Substituting $t = \sqrt{x}$ back,we get $I = 2\tan^{-1}(\sqrt{x}) + A$.
Comparing this with $f(x) + A$,we find $f(x) = 2\tan^{-1}(\sqrt{x})$.

(D) To evaluate the integral $I = \int x \cos(x^2) \, dx$,we use the method of substitution.
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,we get:
$I = \int \cos(t) \cdot \frac{1}{2} \, dt$
$I = \frac{1}{2} \int \cos(t) \, dt$
$I = \frac{1}{2} \sin(t) + c$
Substituting $t = x^2$ back into the expression,we get:
$I = \frac{1}{2} \sin(x^2) + c$.
Thus,the correct option is $D$.

(B) Let $I = \int \frac{x^2 \tan^{-1}(x^3)}{1 + x^6} \, dx$.
Substitute $t = x^3$,then $dt = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} \, dt$.
The integral becomes $I = \frac{1}{3} \int \frac{\tan^{-1}(t)}{1 + t^2} \, dt$.
Now,substitute $z = \tan^{-1}(t)$,then $dz = \frac{1}{1 + t^2} \, dt$.
Substituting these into the integral,we get $I = \frac{1}{3} \int z \, dz$.
Integrating with respect to $z$,we get $I = \frac{1}{3} \cdot \frac{z^2}{2} + c = \frac{z^2}{6} + c$.
Substituting back $z = \tan^{-1}(x^3)$,we get $I = \frac{1}{6}(\tan^{-1}(x^3))^2 + c$.

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

(A) 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,$dt = ({e^x} - ({ - 1}){e^{ - x}}) \,dx = ({e^x} + {e^{ - x}}) \,dx$.
Substituting these into the integral,we get:
$I = \int {\frac{1}{t}} \,dt = \log |t| + c$.
Replacing $t$ with the original expression:
$I = \log |{e^x} - {e^{ - x}}| + c$.

(A) Let $I = \int \frac{\cos x - \sin x}{\sqrt{\sin 2x}} \, dx$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2\sin x \cos x) = 1 - (\sin x - \cos x)^2$.
So,$I = \int \frac{\cos x - \sin x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$.
Let $u = \sin x - \cos x$.
Then $du = (\cos x + \sin x) \, dx$. This substitution does not directly match the numerator.
Let us try $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) \, dx$.
Alternatively,let $u = \sin x - \cos x$. Then $u^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - \sin 2x$.
Thus $\sin 2x = 1 - u^2$.
Then $du = (\cos x + \sin x) \, dx$. This is not the numerator.
Let $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) \, dx$.
Let us re-evaluate: $\sin 2x = -(\sin^2 x + \cos^2 x - 2\sin x \cos x - 1) = -[(\sin x - \cos x)^2 - 1] = 1 - (\sin x - \cos x)^2$.
Let $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) \, dx$.
Actually,the standard substitution is $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) \, dx$.
Wait,the numerator is $\cos x - \sin x$.
Let $u = \sin x + \cos x$. Then $u^2 = 1 + \sin 2x$,so $\sin 2x = u^2 - 1$.
Then $du = (\cos x - \sin x) \, dx$.
Substituting this into the integral: $I = \int \frac{du}{\sqrt{u^2 - 1}} = \ln|u + \sqrt{u^2 - 1}| + c = \ln|\sin x + \cos x + \sqrt{\sin 2x}| + c$.
Given the options,there might be a typo in the question or options. Assuming the integral is $\int \frac{\cos x + \sin x}{\sqrt{\sin 2x}} \, dx$,then $I = \sin^{-1}(\sin x - \cos x) + c$.
If the question is $\int \frac{\cos x - \sin x}{\sqrt{1 - (\sin x + \cos x)^2}} \, dx$,it is invalid.
Based on standard forms,the correct answer is $\sin^{-1}(\sin x - \cos x) + c$.

(A) Let $I = \int {\left( {1 + \frac{1}{{{x^2}}}} \right){e^{x - \frac{1}{x}}}} \,dx$.
Substitute $t = x - \frac{1}{x}$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = 1 - ( - \frac{1}{{{x^2}}}) = 1 + \frac{1}{{{x^2}}}$.
Therefore,$(1 + \frac{1}{{{x^2}}})\,dx = dt$.
Substituting these into the integral,we get $I = \int {{e^t}} \,dt$.
The integral of ${e^t}$ is ${e^t} + c$.
Substituting back $t = x - \frac{1}{x}$,we get $I = {e^{x - \frac{1}{x}}} + c$.

(A) Let $I = \int (x + 3)({x^2} + 6x + 10)^9 \, dx$.
Multiply and divide by $2$:
$I = \frac{1}{2} \int (2x + 6)({x^2} + 6x + 10)^9 \, dx$.
Let $u = x^2 + 6x + 10$.
Then $du = (2x + 6) \, dx$.
Substituting these into the integral:
$I = \frac{1}{2} \int u^9 \, du$.
$I = \frac{1}{2} \cdot \frac{u^{10}}{10} + c$.
$I = \frac{1}{20} u^{10} + c$.
Substituting back $u = x^2 + 6x + 10$:
$I = \frac{1}{20} ({x^2} + 6x + 10)^{10} + c$.

(C) To find the primitive (indefinite integral) of $f(x) = \frac{x}{1 + x^2}$,we evaluate the integral $I = \int \frac{x}{1 + x^2} dx$.
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 $I = \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{1}{t} dt$.
Integrating,we obtain $I = \frac{1}{2} \log_e |t| + C$.
Substituting back $t = 1 + x^2$,we get $I = \frac{1}{2} \log_e(1 + x^2) + C$.

(B) Let $I = \int \frac{1}{x} \log x \, dx$.
Substitute $\log x = t$.
Differentiating both sides with respect to $x$,we get $\frac{1}{x} \, dx = dt$.
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 $t = \log x$,we get $I = \frac{(\log x)^2}{2} + c$.

(C) Let $I = \int \sin^2 x \cos x \, dx$.
Substitute $\sin x = t$.
Differentiating both sides with respect to $x$,we get $\cos x \, dx = dt$.
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 $t = \sin x$,we get $I = \frac{\sin^3 x}{3} + c$.

(B) To evaluate the integral $I = \int e^{x^2} x \, dx$,we use the method of substitution.
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:
$I = \int e^t \cdot \frac{1}{2} dt$
$I = \frac{1}{2} \int e^t \, dt$
$I = \frac{1}{2} e^t + C$
Substituting back $t = x^2$,we get:
$I = \frac{1}{2} e^{x^2} + C$.
Thus,the correct option is $B$.

(C) To solve the integral $I = \int \frac{x^3}{\sqrt{1 + x^4}} \, dx$,we use the method of substitution.
Let $t = 1 + x^4$.
Then,differentiating both sides with respect to $x$,we get $dt = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{1}{4} \, dt$.
Substituting these into the integral:
$I = \int \frac{1}{\sqrt{t}} \cdot \frac{1}{4} \, dt = \frac{1}{4} \int t^{-1/2} \, dt$.
Applying the power rule for integration $\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$:
$I = \frac{1}{4} \cdot \frac{t^{1/2}}{1/2} + c = \frac{1}{4} \cdot 2 \cdot t^{1/2} + c = \frac{1}{2} \sqrt{t} + c$.
Substituting back $t = 1 + x^4$,we get:
$I = \frac{1}{2} \sqrt{1 + x^4} + c$.