Integration by substitution Questions in English

(A) Given the integral $I = \int \frac{dx}{(1 + e^x)(1 + e^{-x})}$.
Rewrite the term $(1 + e^{-x})$ as $(1 + \frac{1}{e^x}) = \frac{e^x + 1}{e^x}$.
Substituting this into the integral,we get $I = \int \frac{dx}{(1 + e^x) \cdot \frac{e^x + 1}{e^x}} = \int \frac{e^x dx}{(1 + e^x)^2}$.
Let $t = 1 + e^x$,then $dt = e^x dx$.
Substituting these into the integral,$I = \int \frac{dt}{t^2} = \int t^{-2} dt$.
Integrating with respect to $t$,we get $I = \frac{t^{-1}}{-1} + C = -\frac{1}{t} + C$.
Substituting back $t = 1 + e^x$,we get $I = -\frac{1}{1 + e^x} + C$.

(C) Let $I = \int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx$.
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:
$I = \int e^t (2 dt) = 2 \int e^t dt$.
Integrating $e^t$ gives $e^t$,so:
$I = 2 e^t + C$.
Substituting back $t = \sqrt{x}$:
$I = 2 e^{\sqrt{x}} + C$.

(D) Let $I = \int \frac{\sin^3 2x}{\cos^5 2x} \, dx$.
We can rewrite the integrand as:
$I = \int \frac{\sin^3 2x}{\cos^3 2x} \cdot \frac{1}{\cos^2 2x} \, dx = \int \tan^3 2x \cdot \sec^2 2x \, dx$.
Now,use the substitution method. Let $u = \tan 2x$.
Then,the derivative is $du = 2 \sec^2 2x \, dx$,which implies $\sec^2 2x \, dx = \frac{du}{2}$.
Substituting these into the integral:
$I = \int u^3 \cdot \frac{du}{2} = \frac{1}{2} \int u^3 \, du$.
Integrating $u^3$ with respect to $u$ gives $\frac{u^4}{4}$.
Thus,$I = \frac{1}{2} \cdot \frac{u^4}{4} + C = \frac{1}{8} u^4 + C$.
Substituting back $u = \tan 2x$,we get:
$I = \frac{1}{8} \tan^4 2x + C$.

(B) Let $I = \int \frac{dx}{(a^2 + x^2)^{3/2}}$.
Substitute $x = a \tan \theta$,then $dx = a \sec^2 \theta \, d\theta$.
Substituting these into the integral:
$I = \int \frac{a \sec^2 \theta}{(a^2 + a^2 \tan^2 \theta)^{3/2}} \, d\theta = \int \frac{a \sec^2 \theta}{(a^2 \sec^2 \theta)^{3/2}} \, d\theta$.
$I = \int \frac{a \sec^2 \theta}{a^3 \sec^3 \theta} \, d\theta = \frac{1}{a^2} \int \frac{1}{\sec \theta} \, d\theta$.
$I = \frac{1}{a^2} \int \cos \theta \, d\theta = \frac{1}{a^2} \sin \theta + C$.
Since $x = a \tan \theta$,we have $\tan \theta = \frac{x}{a}$.
Using the triangle identity,$\sin \theta = \frac{x}{\sqrt{a^2 + x^2}}$.
Therefore,$I = \frac{1}{a^2} \cdot \frac{x}{\sqrt{a^2 + x^2}} + C = \frac{x}{a^2(a^2 + x^2)^{1/2}} + C$.

(C) Let $I = \int {\frac{{{e^{m{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $.
Substitute $t = m{{\tan }^{ - 1}}x$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{m}{1 + x^2}$,which implies $\frac{dx}{1 + x^2} = \frac{dt}{m}$.
Substituting these into the integral:
$I = \int {e^t \cdot \frac{dt}{m}} = \frac{1}{m} \int {e^t dt}$.
Integrating $e^t$ with respect to $t$ gives $e^t + C$.
Therefore,$I = \frac{1}{m} e^t + C = \frac{1}{m} e^{m{{\tan }^{ - 1}}x} + C$.

(D) Let $I = \int \frac{2dx}{\sqrt{1 - 4x^2}}$.
Substitute $2x = t$,then $2dx = dt$.
The integral becomes $I = \int \frac{dt}{\sqrt{1 - t^2}}$.
We know that $\int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$.
Substituting $t = 2x$ back into the expression,we get $I = \sin^{-1}(2x) + c$.

(B) Given integral: $I = \int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}dx}$
Using the property $n\log x = \log(x^n)$ and $e^{\log f(x)} = f(x)$,we have:
$e^{3\log x} = e^{\log(x^3)} = x^3$
So,the integral becomes:
$I = \int \frac{x^3}{x^4 + 1} dx$
Let $u = x^4 + 1$. Then $du = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} du$.
Substituting these into the integral:
$I = \int \frac{1}{u} \cdot \frac{1}{4} du = \frac{1}{4} \int \frac{1}{u} du$
$I = \frac{1}{4} \log|u| + c$
Substituting back $u = x^4 + 1$:
$I = \frac{1}{4} \log(x^4 + 1) + c$

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

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

(C) Let $I = \int \frac{e^x}{e^x + 1} \,dx$.
Substitute $e^x + 1 = t$.
Differentiating both sides with respect to $x$,we get $e^x \,dx = dt$.
Substituting these into the integral,we get $I = \int \frac{1}{t} \,dt$.
The integral of $\frac{1}{t}$ is $\log|t| + c$.
Substituting back $t = e^x + 1$,we get $I = \log|e^x + 1| + c$.
Since $e^x + 1$ is always positive for all real $x$,we can write $I = \log(e^x + 1) + c$.

(D) Let $I = \int \frac{\sin x - \cos x}{\sin x + \cos x} \,dx$.
Substitute $t = \sin x + \cos x$.
Then,$dt = (\cos x - \sin x) \,dx$,which implies $-dt = (\sin x - \cos x) \,dx$.
Substituting these into the integral,we get:
$I = \int \frac{-dt}{t} = -\log |t| + c$.
Using the property of logarithms,$-\log |t| = \log |t^{-1}| = \log \left( \frac{1}{|t|} \right) + c$.
Substituting $t = \sin x + \cos x$ back,we get:
$I = \log \left( \frac{1}{|\sin x + \cos x|} \right) + c$.

(B) Let $t = \tan^{-1} x$.
Then,the derivative is $dt = \frac{1}{1 + x^2} \, dx$.
Substituting these into the integral,we get:
$\int t^3 \, dt = \frac{t^4}{4} + c$.
Substituting back $t = \tan^{-1} x$,the final result is $\frac{(\tan^{-1} x)^4}{4} + c$.

(A) To evaluate the integral $I = \int \frac{\sqrt{x}}{1+x} dx$,substitute $x = t^2$,so $dx = 2t dt$.
Substituting these into the integral:
$I = \int \frac{t}{1+t^2} (2t dt) = 2 \int \frac{t^2}{1+t^2} dt$
$I = 2 \int \frac{t^2+1-1}{1+t^2} dt = 2 \int \left(1 - \frac{1}{1+t^2}\right) dt$
$I = 2 \left( \int 1 dt - \int \frac{1}{1+t^2} dt \right)$
$I = 2(t - \tan^{-1}t) + c$
Substituting $t = \sqrt{x}$ back:
$I = 2(\sqrt{x} - \tan^{-1}\sqrt{x}) + c$.

(D) Let $I = \int \frac{x}{\sqrt{4 - x^4}} \, dx$.
We can rewrite the integral as:
$I = \int \frac{x}{\sqrt{2^2 - (x^2)^2}} \, dx$.
Let $x^2 = t$.
Then,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:
$I = \int \frac{1}{\sqrt{2^2 - t^2}} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int \frac{1}{\sqrt{2^2 - t^2}} \, dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C$,we get:
$I = \frac{1}{2} \sin^{-1} \left( \frac{t}{2} \right) + C$.
Substituting $t = x^2$ back,we get:
$I = \frac{1}{2} \sin^{-1} \left( \frac{x^2}{2} \right) + C$.

(C) Let $I = \int \frac{\sin x}{3 + 4\cos^2 x} dx$.
Substitute $\cos x = t$,then $-\sin x \, dx = dt$,which implies $\sin x \, dx = -dt$.
Substituting these into the integral,we get $I = \int \frac{-dt}{3 + 4t^2}$.
$I = -\frac{1}{4} \int \frac{dt}{t^2 + (\frac{\sqrt{3}}{2})^2}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$,we get:
$I = -\frac{1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1} \left( \frac{t}{\frac{\sqrt{3}}{2}} \right) + c$.
$I = -\frac{1}{4} \cdot \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2t}{\sqrt{3}} \right) + c$.
$I = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2\cos x}{\sqrt{3}} \right) + c$.

(D) Let $I = \int \frac{dx}{\sqrt{x}(x + 9)}$.
Substitute $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.
Substituting these into the integral,we get:
$I = \int \frac{2t \, dt}{t(t^2 + 9)} = 2 \int \frac{dt}{t^2 + 3^2}$.
Using the standard integral formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$,we have:
$I = 2 \cdot \frac{1}{3} \tan^{-1}\left(\frac{t}{3}\right) + C = \frac{2}{3} \tan^{-1}\left(\frac{\sqrt{x}}{3}\right) + C$.

(A) Let $I = \int \frac{\sin 2x}{1 + \cos^2 x} \, dx$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we have:
$I = \int \frac{2 \sin x \cos x}{1 + \cos^2 x} \, dx$.
Let $t = 1 + \cos^2 x$.
Then,$dt = 2 \cos x (-\sin x) \, dx = -\sin 2x \, dx$.
So,$\sin 2x \, dx = -dt$.
Substituting these into the integral:
$I = \int \frac{-dt}{t} = -\log|t| + c$.
Replacing $t$ with $1 + \cos^2 x$:
$I = -\log(1 + \cos^2 x) + c$.

(B) Given $f(x) = \int \frac{x^2 dx}{(1 + x^2)(1 + \sqrt{1 + x^2})}$.
Let $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
Substituting these into the integral:
$f(x) = \int \frac{\tan^2 \theta \cdot \sec^2 \theta \, d\theta}{\sec^2 \theta (1 + \sec \theta)} = \int \frac{\tan^2 \theta \, d\theta}{1 + \sec \theta}$.
Using $\tan^2 \theta = \sec^2 \theta - 1$,we get:
$f(x) = \int \frac{\sec^2 \theta - 1}{1 + \sec \theta} \, d\theta = \int \frac{(\sec \theta - 1)(\sec \theta + 1)}{1 + \sec \theta} \, d\theta = \int (\sec \theta - 1) \, d\theta$.
Integrating,we get $f(x) = \log|\sec \theta + \tan \theta| - \theta + C$.
Since $x = \tan \theta$,then $\sec \theta = \sqrt{1 + x^2}$ and $\theta = \tan^{-1} x$.
Thus,$f(x) = \log|\sqrt{1 + x^2} + x| - \tan^{-1} x + C$.
Given $f(0) = 0$,we have $\log(1) - 0 + C = 0$,which implies $C = 0$.
Therefore,$f(x) = \log(x + \sqrt{1 + x^2}) - \tan^{-1} x$.
For $x = 1$,$f(1) = \log(1 + \sqrt{2}) - \tan^{-1}(1) = \log(1 + \sqrt{2}) - \frac{\pi}{4}$.

(A) Let $I = \int {\sqrt {{e^x} - 1} } dx$.
Substitute ${e^x} - 1 = {t^2}$,which implies ${e^x} = {t^2} + 1$.
Differentiating both sides,we get ${e^x} dx = 2t dt$,so $dx = \frac{2t}{t^2 + 1} dt$.
Substituting these into the integral:
$I = \int t \cdot \frac{2t}{t^2 + 1} dt = \int \frac{2t^2}{t^2 + 1} dt$.
Now,rewrite the integrand:
$I = \int \frac{2(t^2 + 1) - 2}{t^2 + 1} dt = \int \left( 2 - \frac{2}{t^2 + 1} \right) dt$.
Integrating term by term:
$I = 2t - 2 \tan^{-1}(t) + c$.
Substituting back $t = \sqrt{e^x - 1}$:
$I = 2\sqrt{e^x - 1} - 2 \tan^{-1}(\sqrt{e^x - 1}) + c = 2\left[ \sqrt{e^x - 1} - \tan^{-1}(\sqrt{e^x - 1}) \right] + c$.

(C) Let $I = \int \frac{\sin \theta + \cos \theta}{\sqrt{\sin 2\theta}} d\theta$.
We know that $\sin 2\theta = 1 - (1 - 2\sin \theta \cos \theta) = 1 - (\sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta) = 1 - (\sin \theta - \cos \theta)^2$.
Substituting this into the integral,we get:
$I = \int \frac{\sin \theta + \cos \theta}{\sqrt{1 - (\sin \theta - \cos \theta)^2}} d\theta$.
Let $t = \sin \theta - \cos \theta$.
Then $dt = (\cos \theta + \sin \theta) d\theta$.
Substituting $t$ and $dt$ into the integral:
$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$.
Replacing $t$ with $(\sin \theta - \cos \theta)$,we get:
$I = \sin^{-1}(\sin \theta - \cos \theta) + c$.

(A) Let $I = \int \sin^3 x \cos^2 x \, dx$.
We can write $\sin^3 x$ as $\sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x$.
So,$I = \int (1 - \cos^2 x) \cos^2 x \sin x \, dx$.
Let $\cos x = t$. Then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Substituting these into the integral,we get:
$I = \int (1 - t^2) t^2 (-dt) = \int (t^4 - t^2) \, dt$.
Integrating with respect to $t$:
$I = \frac{t^5}{5} - \frac{t^3}{3} + c$.
Substituting back $t = \cos x$:
$I = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + c$.

(A) Let $I = \int \frac{x}{x^4 - 1} dx$.
Substitute $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \frac{1}{2} \int \frac{1}{t^2 - 1} dt$.
Using the standard formula $\int \frac{1}{t^2 - a^2} dt = \frac{1}{2a} \log \left| \frac{t - a}{t + a} \right| + c$:
$I = \frac{1}{2} \left( \frac{1}{2(1)} \log \left| \frac{t - 1}{t + 1} \right| \right) + c$.
$I = \frac{1}{4} \log \left| \frac{x^2 - 1}{x^2 + 1} \right| + c$.

(A) Let $I = \int \sin^5 x \cos^4 x \, dx$.
Substitute $\cos x = t$,then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
We can write $\sin^5 x$ as $(\sin^2 x)^2 \sin x = (1 - \cos^2 x)^2 \sin x$.
Substituting these into the integral:
$I = \int (1 - t^2)^2 t^4 (-dt) = - \int (1 - 2t^2 + t^4) t^4 \, dt$
$I = - \int (t^4 - 2t^6 + t^8) \, dt$
$I = - (\frac{t^5}{5} - \frac{2t^7}{7} + \frac{t^9}{9}) + c$
$I = - \frac{t^5}{5} + \frac{2t^7}{7} - \frac{t^9}{9} + c$
Substituting $t = \cos x$ back:
$I = - \frac{1}{5}\cos^5 x + \frac{2}{7}\cos^7 x - \frac{1}{9}\cos^9 x + c$.

(B) To evaluate the integral $I = \int \frac{dx}{\sqrt{2x - x^2}}$,we first complete the square for the expression inside the square root:
$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x - 1)^2$.
Substituting this back into the integral,we get:
$I = \int \frac{dx}{\sqrt{1 - (x - 1)^2}}$.
Using the standard integration formula $\int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + c$,where $u = x - 1$ and $du = dx$,we obtain:
$I = \sin^{-1}(x - 1) + c$.

(C) To evaluate the integral $I = \int \frac{dx}{5 + 4\cos x}$,we use the substitution $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$.
Substituting this into the integral,we get:
$I = \int \frac{dx}{5 + 4\left[ \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \right]} = \int \frac{\sec^2(x/2) dx}{5(1 + \tan^2(x/2)) + 4(1 - \tan^2(x/2))}$
$I = \int \frac{\sec^2(x/2) dx}{5 + 5\tan^2(x/2) + 4 - 4\tan^2(x/2)} = \int \frac{\sec^2(x/2) dx}{9 + \tan^2(x/2)}$
Let $t = \tan(x/2)$,then $dt = \frac{1}{2}\sec^2(x/2) dx$,which implies $2 dt = \sec^2(x/2) dx$.
Substituting these into the integral:
$I = \int \frac{2 dt}{3^2 + t^2} = 2 \cdot \frac{1}{3} \tan^{-1}\left( \frac{t}{3} \right) + c = \frac{2}{3} \tan^{-1}\left( \frac{1}{3} \tan \frac{x}{2} \right) + c$.

(B) To evaluate the integral $I = \int \frac{dx}{1 + 3\sin^2 x}$,we divide the numerator and denominator by $\cos^2 x$ or rewrite the denominator using the identity $1 = \sin^2 x + \cos^2 x$.
$I = \int \frac{dx}{\sin^2 x + \cos^2 x + 3\sin^2 x} = \int \frac{dx}{4\sin^2 x + \cos^2 x}$.
Divide numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x dx}{4\tan^2 x + 1} = \frac{1}{4} \int \frac{\sec^2 x dx}{\tan^2 x + \frac{1}{4}}$.
Let $t = \tan x$,then $dt = \sec^2 x dx$.
$I = \frac{1}{4} \int \frac{dt}{t^2 + (\frac{1}{2})^2}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{4} \cdot \frac{1}{1/2} \tan^{-1}(\frac{t}{1/2}) + c = \frac{1}{4} \cdot 2 \tan^{-1}(2t) + c = \frac{1}{2} \tan^{-1}(2\tan x) + c$.

(A) Let $I = \int \frac{dx}{7 + 5\cos x}$.
Using the substitution $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$,we get:
$I = \int \frac{dx}{7 + 5 \left( \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \right)} = \int \frac{(1 + \tan^2(x/2)) dx}{7(1 + \tan^2(x/2)) + 5(1 - \tan^2(x/2))}$
$I = \int \frac{\sec^2(x/2) dx}{7 + 7\tan^2(x/2) + 5 - 5\tan^2(x/2)} = \int \frac{\sec^2(x/2) dx}{12 + 2\tan^2(x/2)}$
$I = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{6 + \tan^2(x/2)}$
Let $\tan(x/2) = t$,then $\frac{1}{2} \sec^2(x/2) dx = dt$.
$I = \int \frac{dt}{t^2 + (\sqrt{6})^2} = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{t}{\sqrt{6}} \right) + c$
Substituting back $t = \tan(x/2)$:
$I = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{\tan(x/2)}{\sqrt{6}} \right) + c$.

(A) Let $\log x = t$. Then,$\frac{1}{x} dx = dt$.
Substituting this into the integral,we get:
$\int \frac{dt}{t^2 + 4t - 1}$.
Completing the square in the denominator:
$t^2 + 4t - 1 = (t^2 + 4t + 4) - 4 - 1 = (t + 2)^2 - 5 = (t + 2)^2 - (\sqrt{5})^2$.
Using the standard integral formula $\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \log \left| \frac{u - a}{u + a} \right| + c$:
$\int \frac{dt}{(t + 2)^2 - (\sqrt{5})^2} = \frac{1}{2\sqrt{5}} \log \left| \frac{t + 2 - \sqrt{5}}{t + 2 + \sqrt{5}} \right| + c$.
Substituting $t = \log x$ back into the expression:
$\frac{1}{2\sqrt{5}} \log \left| \frac{\log x + 2 - \sqrt{5}}{\log x + 2 + \sqrt{5}} \right| + c$.

(B) Given integral is $I = \int \frac{dx}{x(x^7 + 1)}$.
Multiply the numerator and denominator by $x^6$ to get $I = \int \frac{x^6 dx}{x^7(x^7 + 1)}$.
Let $x^7 = u$,then $7x^6 dx = du$,which implies $x^6 dx = \frac{du}{7}$.
Substituting these into the integral,we get $I = \int \frac{du/7}{u(u + 1)} = \frac{1}{7} \int \frac{du}{u(u + 1)}$.
Using partial fractions,$\frac{1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}$.
Thus,$I = \frac{1}{7} \int \left( \frac{1}{u} - \frac{1}{u + 1} \right) du = \frac{1}{7} (\log |u| - \log |u + 1|) + c$.
$I = \frac{1}{7} \log \left| \frac{u}{u + 1} \right| + c$.
Substituting $u = x^7$ back,we get $I = \frac{1}{7} \log \left( \frac{x^7}{x^7 + 1} \right) + c$.

(A) Let $I = \int \frac{dx}{(1 + x^2)\sqrt{p^2 + q^2(\tan^{-1}x)^2}}$.
Substitute $t = q\tan^{-1}x$.
Then $dt = \frac{q}{1 + x^2} dx$,which implies $\frac{dx}{1 + x^2} = \frac{dt}{q}$.
Substituting these into the integral,we get:
$I = \int \frac{1}{\sqrt{p^2 + t^2}} \cdot \frac{dt}{q} = \frac{1}{q} \int \frac{dt}{\sqrt{p^2 + t^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log |x + \sqrt{x^2 + a^2}| + c$,we have:
$I = \frac{1}{q} \log |t + \sqrt{t^2 + p^2}| + c$.
Substituting $t = q\tan^{-1}x$ back into the expression:
$I = \frac{1}{q} \log |q\tan^{-1}x + \sqrt{q^2(\tan^{-1}x)^2 + p^2}| + c$.

(C) Let $I = \int \frac{x^5}{\sqrt{1 + x^3}} dx$.
Put $1 + x^3 = t^2$,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:
$I = \int \frac{x^3 \cdot x^2 dx}{\sqrt{1 + x^3}} = \int \frac{(t^2 - 1) \cdot \frac{2}{3} t dt}{t}$
$I = \frac{2}{3} \int (t^2 - 1) dt$
$I = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + c$
$I = \frac{2}{9} t^3 - \frac{2}{3} t + c$
Since $t = (1 + x^3)^{1/2}$,we have:
$I = \frac{2}{9} (1 + x^3)^{3/2} - \frac{2}{3} (1 + x^3)^{1/2} + c$.

(A) Let $I = \int \frac{a \, dx}{b + c e^x}$.
Multiply numerator and denominator by $e^{-x}$:
$I = \int \frac{a e^{-x} \, dx}{b e^{-x} + c}$.
Let $u = b e^{-x} + c$,then $du = -b e^{-x} \, dx$,which implies $e^{-x} \, dx = -\frac{du}{b}$.
Substituting these into the integral:
$I = a \int \frac{-du/b}{u} = -\frac{a}{b} \int \frac{du}{u} = -\frac{a}{b} \log |u| + C$.
Substituting back $u = b e^{-x} + c = \frac{b + c e^x}{e^x}$:
$I = -\frac{a}{b} \log \left| \frac{b + c e^x}{e^x} \right| + C = \frac{a}{b} \log \left| \frac{e^x}{b + c e^x} \right| + C$.

(B) Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.
Substituting these into the integral:
$\int \sin \sqrt{x} \, dx = \int \sin(t) \cdot 2t \, dt = 2 \int t \sin(t) \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = \sin(t) \, dt$:
$du = dt$ and $v = -\cos(t)$.
So,$2 \int t \sin(t) \, dt = 2 [t(-\cos(t)) - \int (-\cos(t)) \, dt] = 2 [-t \cos(t) + \sin(t)] + c$.
Substituting $t = \sqrt{x}$ back:
$= 2[\sin \sqrt{x} - \sqrt{x} \cos \sqrt{x}] + c$.

(A) Let $x = 3\sin \theta$,then $dx = 3\cos \theta \, d\theta$.
Substituting these into the integral:
$\int \frac{9\sin^2 \theta}{(9 - 9\sin^2 \theta)^{3/2}} \cdot 3\cos \theta \, d\theta = \int \frac{27\sin^2 \theta \cos \theta}{(9\cos^2 \theta)^{3/2}} \, d\theta$
$= \int \frac{27\sin^2 \theta \cos \theta}{27\cos^3 \theta} \, d\theta = \int \tan^2 \theta \, d\theta$
$= \int (\sec^2 \theta - 1) \, d\theta = \tan \theta - \theta + c$
Since $\sin \theta = \frac{x}{3}$,we have $\theta = \sin^{-1}\left(\frac{x}{3}\right)$ and $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x/3}{\sqrt{1 - (x/3)^2}} = \frac{x}{\sqrt{9 - x^2}}$.
Thus,the integral is $\frac{x}{\sqrt{9 - x^2}} - \sin^{-1}\left(\frac{x}{3}\right) + c$.

(C) Divide the numerator and denominator by $x^2$:
$\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} - 1 + \frac{1}{{{x^2}}}}}\,dx} $
$= \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 1}}\,dx} $
Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{{{x^2}}})\,dx$.
Substituting these into the integral:
$\int {\frac{{dt}}{{{t^2} + 1}} = {{\tan }^{ - 1}}t + c} $
Substituting back $t = x - \frac{1}{x}$:
$= {\tan ^{ - 1}}\left( {x - \frac{1}{x}} \right) + c = {\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c$.

(A) Let $I = \int {\frac{{\sqrt {{x^2} - {a^2}} }}{x}dx}$.
Substitute $\sqrt {{x^2} - {a^2}} = t$,then ${x^2} - {a^2} = {t^2}$,which implies ${x^2} = {a^2} + {t^2}$.
Differentiating both sides,$2xdx = 2tdt$,so $xdx = tdt$.
Now,multiply the numerator and denominator by $x$:
$I = \int {\frac{{\sqrt {{x^2} - {a^2}} \cdot x}}{{{x^2}}}dx} = \int {\frac{t}{{{a^2} + {t^2}}} \cdot tdt} = \int {\frac{{{t^2}}}{{{a^2} + {t^2}}}dt}$.
$I = \int {\left( {\frac{{{t^2} + {a^2} - {a^2}}}{{{a^2} + {t^2}}}} \right)dt} = \int {\left( {1 - \frac{{{a^2}}}{{{a^2} + {t^2}}}} \right)dt}$.
$I = \int {1dt} - {a^2}\int {\frac{1}{{{a^2} + {t^2}}}dt} = t - {a^2} \cdot \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{t}{a}} \right) + C$.
$I = t - a{\tan ^{ - 1}}\left( {\frac{t}{a}} \right) + C$.
Substituting back $t = \sqrt {{x^2} - {a^2}}$,we get:
$I = \sqrt {{x^2} - {a^2}} - a{\tan ^{ - 1}}\left( {\frac{{\sqrt {{x^2} - {a^2}} }}{a}} \right) + C$.

(A) Let $I = \int \tan^3 2x \sec 2x \, dx$.
We can rewrite $\tan^3 2x$ as $\tan^2 2x \cdot \tan 2x$.
Since $\tan^2 2x = \sec^2 2x - 1$,the integral becomes:
$I = \int (\sec^2 2x - 1) \sec 2x \tan 2x \, dx$.
Let $u = \sec 2x$. Then $du = 2 \sec 2x \tan 2x \, dx$,which implies $\sec 2x \tan 2x \, dx = \frac{1}{2} du$.
Substituting these into the integral:
$I = \int (u^2 - 1) \cdot \frac{1}{2} du$
$I = \frac{1}{2} \int (u^2 - 1) \, du$
$I = \frac{1}{2} \left( \frac{u^3}{3} - u \right) + c$
$I = \frac{u^3}{6} - \frac{u}{2} + c$
Substituting $u = \sec 2x$ back into the equation:
$I = \frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + c$.

(A) Let $I = \int \sqrt{\frac{a-x}{x}} \, dx$.
Substitute $x = a \sin^2 \theta$,then $dx = 2a \sin \theta \cos \theta \, d\theta$.
$I = \int \sqrt{\frac{a - a \sin^2 \theta}{a \sin^2 \theta}} \cdot (2a \sin \theta \cos \theta) \, d\theta$
$I = \int \frac{\cos \theta}{\sin \theta} \cdot 2a \sin \theta \cos \theta \, d\theta = 2a \int \cos^2 \theta \, d\theta$
$I = a \int (1 + \cos 2\theta) \, d\theta = a(\theta + \frac{\sin 2\theta}{2}) + c = a(\theta + \sin \theta \cos \theta) + c$
Since $x = a \sin^2 \theta$,we have $\sin \theta = \sqrt{\frac{x}{a}}$ and $\theta = \sin^{-1} \sqrt{\frac{x}{a}}$.
Also,$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{x}{a}} = \sqrt{\frac{a-x}{a}}$.
Substituting these back,$I = a \left[ \sin^{-1} \sqrt{\frac{x}{a}} + \sqrt{\frac{x}{a}} \sqrt{\frac{a-x}{a}} \right] + c$.

(A) Given the integral $I = \int \frac{\sin x - \cos x}{\sqrt{1 - \sin 2x}} e^{\sin x} \cos x \, dx$.
We know that $1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\sin x - \cos x)^2$.
Thus,$\sqrt{1 - \sin 2x} = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|$.
Since $x \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$,we have $\sin x > \cos x$,so $|\sin x - \cos x| = \sin x - \cos x$.
Substituting this into the integral:
$I = \int \frac{\sin x - \cos x}{\sin x - \cos x} e^{\sin x} \cos x \, dx = \int e^{\sin x} \cos x \, dx$.
Let $u = \sin x$,then $du = \cos x \, dx$.
$I = \int e^u \, du = e^u + c = e^{\sin x} + c$.

(B) Let $I = \int \frac{x}{x^4 + x^2 + 1} dx$.
Substitute $u = x^2$,then $du = 2x dx$,or $x dx = \frac{1}{2} du$.
The integral becomes $I = \frac{1}{2} \int \frac{du}{u^2 + u + 1}$.
Complete the square in the denominator: $u^2 + u + 1 = (u + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1} \left( \frac{u + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C$.
$I = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2u + 1}{\sqrt{3}} \right) + C$.
Substituting $u = x^2$ back,we get $I = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2x^2 + 1}{\sqrt{3}} \right) + C$.

(A) To evaluate ${I_1} = \int_e^{{e^2}} \frac{dx}{\log x}$,we use the substitution method.
Let $\log x = u$. Then $x = e^u$,which implies $dx = e^u du$.
When $x = e$,$u = \log e = 1$.
When $x = e^2$,$u = \log e^2 = 2$.
Substituting these into the integral ${I_1}$,we get:
${I_1} = \int_1^2 \frac{e^u}{u} du$.
Since the variable of integration is a dummy variable,we can replace $u$ with $x$:
${I_1} = \int_1^2 \frac{e^x}{x} dx$.
Comparing this with ${I_2} = \int_1^2 \frac{e^x}{x} dx$,we see that ${I_1} = {I_2}$.

(A) Let $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$.
Factor the denominator:
$\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x = \sin^2 x(\sin^3 x + \cos^3 x) + \cos^2 x(\sin^3 x + \cos^3 x) = (\sin^2 x + \cos^2 x)(\sin^3 x + \cos^3 x) = (\sin^3 x + \cos^3 x)$.
Thus,$I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$.
Divide numerator and denominator by $\cos^6 x$:
$I = \int \frac{\tan^2 x \sec^2 x}{(\tan^3 x + 1)^2} dx$.
Let $t = 1 + \tan^3 x$,then $dt = 3 \tan^2 x \sec^2 x dx$,so $\tan^2 x \sec^2 x dx = \frac{dt}{3}$.
$I = \frac{1}{3} \int \frac{dt}{t^2} = \frac{1}{3} (-t^{-1}) + C = \frac{-1}{3(1 + \tan^3 x)} + C$.

(C) Assertion $(A)$ is true because the integral $\int \frac{1}{x^2 + a^2} dx$ can be solved by substituting $x = a \tan \theta$,which gives $dx = a \sec^2 \theta d\theta$. Substituting these into the integral yields $\int \frac{a \sec^2 \theta}{a^2 \tan^2 \theta + a^2} d\theta = \int \frac{a \sec^2 \theta}{a^2 \sec^2 \theta} d\theta = \frac{1}{a} \int d\theta = \frac{1}{a} \theta + C = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$.
Reason $(R)$ is false because not all integrands can be integrated by the method of substitution. There are many other methods such as integration by parts,partial fractions,and direct standard formulas.
Therefore,$A$ is true but $R$ is false.

(B) Let $I = \int {{e^{{{\cos }^2}x}}\sin 2x\,dx}$.
Substitute $t = {\cos ^2}x$.
Then,differentiating both sides with respect to $x$,we get $dt = 2\cos x(-\sin x)\,dx = -\sin 2x\,dx$.
This implies $\sin 2x\,dx = -dt$.
Substituting these into the integral,we get $I = \int {e^t(-dt)} = -\int {e^t\,dt}$.
Integrating $e^t$,we get $I = -{e^t} + c$.
Substituting back $t = {\cos ^2}x$,we get $I = -{e^{{{\cos }^2}x}} + c$.

(B) We have the integral $I = \int \frac{1}{x^2(x^4 + 1)^{3/4}} dx$.
Taking $x^4$ common from the bracket,we get:
$I = \int \frac{1}{x^2 [x^4(1 + \frac{1}{x^4})]^{3/4}} dx = \int \frac{1}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} dx = \int \frac{1}{x^5 (1 + \frac{1}{x^4})^{3/4}} dx$.
Let $t = 1 + \frac{1}{x^4}$.
Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dx}{x^5} = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$.
Substituting $t$ back:
$I = -(1 + \frac{1}{x^4})^{1/4} + c = -(\frac{x^4 + 1}{x^4})^{1/4} + c = -\frac{(x^4 + 1)^{1/4}}{x} + c$.

(D) Let $I = \int \sqrt{2 + \sin 3x} \cdot \cos 3x \, dx$.
Substitute $t = 2 + \sin 3x$.
Then,differentiating with respect to $x$,we get $dt = 3 \cos 3x \, dx$,which implies $\cos 3x \, dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \sqrt{t} \cdot \frac{1}{3} dt = \frac{1}{3} \int t^{1/2} dt$.
Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + c$:
$I = \frac{1}{3} \cdot \frac{t^{3/2}}{3/2} + c = \frac{1}{3} \cdot \frac{2}{3} t^{3/2} + c = \frac{2}{9} t^{3/2} + c$.
Substituting back $t = 2 + \sin 3x$:
$I = \frac{2}{9}(2 + \sin 3x)^{3/2} + c$.

(B) Let $I = \int {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$.
We can rewrite the integrand as:
$I = \int {\frac{{\sqrt {{x^2}(1 + 1/x^2)} [\log (x^2(1 + 1/x^2))]}}{{{x^4}}}} dx = \int {\frac{{x\sqrt {1 + 1/x^2} [\log x^2 + \log(1 + 1/x^2) - 2\log x]}}{{{x^4}}}} dx$.
Since $\log x^2 = 2\log x$,the expression simplifies to:
$I = \int {\sqrt {1 + \frac{1}{{{x^2}}}} \log \left( {1 + \frac{1}{{{x^2}}}} \right) \cdot \frac{1}{{{x^3}}}} dx$.
Let $1 + \frac{1}{{{x^2}}} = t$. Then $-\frac{2}{{{x^3}}} dx = dt$,or $\frac{1}{{{x^3}}} dx = -\frac{1}{2} dt$.
Substituting these into the integral:
$I = - \frac{1}{2} \int {\sqrt t \log t} dt$.
Using integration by parts,$\int u dv = uv - \int v du$,with $u = \log t$ and $dv = \sqrt t dt$:
$I = - \frac{1}{2} \left[ \log t \cdot \frac{t^{3/2}}{3/2} - \int \frac{1}{t} \cdot \frac{t^{3/2}}{3/2} dt \right] = - \frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \int t^{1/2} dt \right]$.
$I = - \frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \cdot \frac{t^{3/2}}{3/2} \right] + c = - \frac{1}{3} t^{3/2} \log t + \frac{2}{9} t^{3/2} + c$.
$I = - \frac{1}{3} t^{3/2} \left[ \log t - \frac{2}{3} \right] + c$.
Substituting $t = 1 + \frac{1}{{{x^2}}}$ back:
$I = - \frac{1}{3} {\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}} \left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + c$.

(C) Let $I = \int \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}} \, dx = \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx$.
Since $\sqrt{\sin^2 x} = \sin x$ (assuming $\sin x > 0$),we have $I = \int \sqrt{\frac{\cos x}{1 - \cos^3 x}} \sin x \, dx$.
Let $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$.
Substituting these into the integral: $I = - \int \sqrt{\frac{t}{1 - t^3}} \, dt = - \int \frac{\sqrt{t}}{\sqrt{1 - (t^{3/2})^2}} \, dt$.
Let $u = t^{3/2}$,then $du = \frac{3}{2} t^{1/2} \, dt = \frac{3}{2} \sqrt{t} \, dt$,which implies $\sqrt{t} \, dt = \frac{2}{3} du$.
Thus,$I = - \int \frac{2/3}{\sqrt{1 - u^2}} \, du = - \frac{2}{3} \sin^{-1}(u) + c = - \frac{2}{3} \sin^{-1}(t^{3/2}) + c$.
Using the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,we have $-\frac{2}{3} \sin^{-1}(t^{3/2}) + c = \frac{2}{3} \cos^{-1}(t^{3/2}) + K$.
Substituting $t = \cos x$,we get $I = \frac{2}{3} \cos^{-1}(\cos^{3/2} x) + c$.

(A) Let $I = \int \frac{1}{x \cdot l^1(x) \cdot l^2(x) \cdot ... \cdot l^r(x)} \, dx$.
Consider the substitution $t = l^{r+1}(x)$.
By the chain rule,the derivative of $l^{r+1}(x) = \log(l^r(x))$ is $\frac{d}{dx} l^{r+1}(x) = \frac{1}{l^r(x)} \cdot \frac{d}{dx} l^r(x)$.
Continuing this process,we find that $\frac{d}{dx} l^{r+1}(x) = \frac{1}{l^r(x) \cdot l^{r-1}(x) \cdot ... \cdot l^1(x) \cdot x}$.
Thus,$dt = \frac{1}{x \cdot l^1(x) \cdot ... \cdot l^r(x)} \, dx$.
Substituting this into the integral,we get $\int 1 \, dt = t + c$.
Therefore,the integral is $l^{r+1}(x) + c$.

(B) Let $I = \int {{2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}dx}$.
Substitute $u = 2^x$,then $du = 2^x \ln 2 \, dx$,so $2^x \, dx = \frac{du}{\ln 2}$.
Now,let $v = 2^u = 2^{2^x}$,then $dv = 2^u \ln 2 \, du = 2^{2^x} \ln 2 \, du$,so $2^{2^x} \, du = \frac{dv}{\ln 2}$.
Finally,let $t = 2^v = 2^{2^{2^x}}$,then $dt = 2^v \ln 2 \, dv = 2^{2^{2^x}} \ln 2 \, dv$,so $2^{2^{2^x}} \, dv = \frac{dt}{\ln 2}$.
Alternatively,using substitution $t = 2^{2^{2^x}}$,we differentiate with respect to $x$:
$\frac{dt}{dx} = 2^{2^{2^x}} \cdot \ln 2 \cdot \frac{d}{dx}(2^{2^x}) = 2^{2^{2^x}} \cdot \ln 2 \cdot (2^{2^x} \cdot \ln 2 \cdot \frac{d}{dx}(2^x)) = 2^{2^{2^x}} \cdot 2^{2^x} \cdot 2^x \cdot (\ln 2)^3$.
Thus,$2^{2^{2^x}} \cdot 2^{2^x} \cdot 2^x \, dx = \frac{dt}{(\ln 2)^3}$.
Therefore,$\int \frac{dt}{(\ln 2)^3} = \frac{t}{(\ln 2)^3} + c = \frac{2^{2^{2^x}}}{(\ln 2)^3} + c$.