Integration by substitution Questions in English

(B) Let $I = \int \frac{\cos x}{\sqrt{4 \sin ^2 x+4 \sin x+5}} d x$.
Substitute $\sin x = t$,then $\cos x d x = d t$.
Thus,$I = \int \frac{d t}{\sqrt{4 t^2+4 t+5}} = \frac{1}{2} \int \frac{d t}{\sqrt{t^2+t+\frac{5}{4}}}$.
Completing the square in the denominator: $t^2+t+\frac{5}{4} = (t+\frac{1}{2})^2 + 1$.
So,$I = \frac{1}{2} \int \frac{d t}{\sqrt{(t+\frac{1}{2})^2 + 1}}$.
Using the formula $\int \frac{du}{\sqrt{u^2+a^2}} = \sinh^{-1}(\frac{u}{a}) + C$,we get:
$I = \frac{1}{2} \sinh^{-1}(t+\frac{1}{2}) + C$.
Substituting $t = \sin x$,we have $I = \frac{1}{2} \sinh^{-1}(\sin x + \frac{1}{2}) + C$.
Comparing this with $\frac{1}{2} \sinh^{-1}(f(x)) + C$,we get $f(x) = \sin x + \frac{1}{2}$.
Therefore,$2 f(x) = 2(\sin x + \frac{1}{2}) = 2 \sin x + 1$.

(B) Let $I = \int \frac{x^2}{(\sqrt{4-x^2})^3} dx$.
Substitute $x = 2 \sin \theta$,then $dx = 2 \cos \theta d\theta$.
Substituting these into the integral:
$I = \int \frac{(2 \sin \theta)^2}{(\sqrt{4-4 \sin^2 \theta})^3} (2 \cos \theta) d\theta$
$I = \int \frac{4 \sin^2 \theta \cdot 2 \cos \theta}{(2 \cos \theta)^3} d\theta$
$I = \int \frac{8 \sin^2 \theta \cos \theta}{8 \cos^3 \theta} d\theta = \int \tan^2 \theta d\theta$
$I = \int (\sec^2 \theta - 1) d\theta = \tan \theta - \theta + C$.
Since $x = 2 \sin \theta$,we have $\sin \theta = \frac{x}{2}$.
Then $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x/2}{\sqrt{1-(x/2)^2}} = \frac{x}{\sqrt{4-x^2}}$.
Also,$\theta = \sin^{-1}(\frac{x}{2}) = \tan^{-1}(\frac{x}{\sqrt{4-x^2}})$.
Thus,$I = \frac{x}{\sqrt{4-x^2}} - \tan^{-1}(\frac{x}{\sqrt{4-x^2}}) + C$.

(A) Let $I = \int \left( \frac{x^4-x}{x^{20}} \right)^{1/4} dx$.
$I = \int \left( \frac{x^4(1 - 1/x^3)}{x^{20}} \right)^{1/4} dx = \int \frac{1}{x^4} (1 - x^{-3})^{1/4} dx$.
Let $1 - x^{-3} = t$.
Then,differentiating both sides with respect to $x$,we get $3x^{-4} dx = dt$,which implies $\frac{1}{x^4} dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int (t)^{1/4} \cdot \frac{1}{3} dt = \frac{1}{3} \cdot \frac{t^{5/4}}{5/4} + C = \frac{4}{15} t^{5/4} + C$.
Substituting $t = 1 - \frac{1}{x^3} = \frac{x^3-1}{x^3}$ back:
$I = \frac{4}{15} \left( \frac{x^3-1}{x^3} \right)^{5/4} + C = \frac{4}{15} \left( \frac{(x^3-1)^5}{x^{15}} \right)^{1/4} + C$.

(D) Let $I = \int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x$.
Substitute $t = x + \sqrt{1+x^2}$.
Differentiating with respect to $x$,we get $dt = \left(1 + \frac{2x}{2\sqrt{1+x^2}}\right) dx = \left(1 + \frac{x}{\sqrt{1+x^2}}\right) dx = \left(\frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}\right) dx$.
Thus,the integral becomes $I = \int t dt$.
Integrating $t$ with respect to $t$,we get $I = \frac{t^2}{2} + C$.
Substituting $t$ back,we get $I = \frac{\left(x+\sqrt{1+x^2}\right)^2}{2} + C$.

(D) We have,$I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} dx$.
Divide the numerator and denominator inside the square root by $x^4$:
$I = \int \frac{x^2-1}{x^3 \cdot x^2 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx = \int \frac{x^2-1}{x^5 \sqrt{2 - 2x^{-2} + x^{-4}}} dx$.
$I = \int \frac{x^{-3} - x^{-5}}{\sqrt{2 - 2x^{-2} + x^{-4}}} dx$.
Let $t = \sqrt{2 - 2x^{-2} + x^{-4}}$.
Then $t^2 = 2 - 2x^{-2} + x^{-4}$.
Differentiating both sides with respect to $x$:
$2t \frac{dt}{dx} = 4x^{-3} - 4x^{-5} = 4(x^{-3} - x^{-5})$.
So,$(x^{-3} - x^{-5}) dx = \frac{1}{2} t dt$.
Substituting this into the integral:
$I = \int \frac{\frac{1}{2} t dt}{t} = \frac{1}{2} \int dt = \frac{1}{2} t + c$.
Substituting back $t$:
$I = \frac{1}{2} \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + c$.

(B) Let $I = \int \frac{d x}{\left(e^x+e^{-x}\right)^2} = \int \frac{d x}{\left(e^x+\frac{1}{e^x}\right)^2}$
$= \int \frac{e^{2 x} d x}{\left(e^{2 x}+1\right)^2}$
Let $t = e^{2 x}+1$,then $dt = 2e^{2 x} dx$,which implies $e^{2 x} dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{1}{2} \frac{dt}{t^2} = \frac{1}{2} \int t^{-2} dt$
$= \frac{1}{2} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{2t} + c$
Substituting $t = e^{2 x}+1$ back:
$I = -\frac{1}{2\left(e^{2 x}+1\right)} + c$

(A) Let $I = \int \frac{3^x}{\sqrt{9^x-1}} dx$.
Substitute $3^x = t$.
Then,differentiating both sides with respect to $x$,we get $3^x \log 3 dx = dt$,which implies $3^x dx = \frac{dt}{\log 3}$.
Substituting these into the integral,we get:
$I = \frac{1}{\log 3} \int \frac{dt}{\sqrt{t^2-1}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2-a^2}} = \log \left|x + \sqrt{x^2-a^2}\right| + c$,we have:
$I = \frac{1}{\log 3} \log \left|t + \sqrt{t^2-1}\right| + c$.
Replacing $t$ with $3^x$,we get:
$I = \frac{1}{\log 3} \log \left|3^x + \sqrt{9^x-1}\right| + c$.

(D) Let $I = \int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) dx$.
Substitute $x = a \tan^2 \theta$,so $dx = 2a \tan \theta \sec^2 \theta d\theta$.
Then $\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a \tan^2 \theta}{a(1+\tan^2 \theta)}} = \sqrt{\sin^2 \theta} = \sin \theta$.
Thus,$I = \int \sin^{-1}(\sin \theta) \cdot 2a \tan \theta \sec^2 \theta d\theta = 2a \int \theta \tan \theta \sec^2 \theta d\theta$.
Using integration by parts,let $u = \theta$ and $dv = \tan \theta \sec^2 \theta d\theta$. Then $du = d\theta$ and $v = \frac{\tan^2 \theta}{2}$.
$I = 2a \left[ \theta \cdot \frac{\tan^2 \theta}{2} - \int \frac{\tan^2 \theta}{2} d\theta \right] = a \theta \tan^2 \theta - a \int (\sec^2 \theta - 1) d\theta$.
$I = a \theta \tan^2 \theta - a(\tan \theta - \theta) + C = a \theta (\tan^2 \theta + 1) - a \tan \theta + C = a \theta \sec^2 \theta - a \tan \theta + C$.
Since $\tan^2 \theta = \frac{x}{a}$,we have $\theta = \tan^{-1} \sqrt{\frac{x}{a}}$ and $\sec^2 \theta = 1 + \frac{x}{a} = \frac{a+x}{a}$.
Substituting back: $I = a \left( \tan^{-1} \sqrt{\frac{x}{a}} \right) \left( \frac{a+x}{a} \right) - a \sqrt{\frac{x}{a}} + C = (a+x) \tan^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + C$.
Therefore,$A(x) = (a+x) \tan^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax}$.

(B) Let $I = \int \sqrt{e^x-4} \, dx$.
Put $e^x-4 = t^2$.
Then $e^x \, dx = 2t \, dt$,which implies $dx = \frac{2t}{e^x} \, dt = \frac{2t}{t^2+4} \, dt$.
Substituting these into the integral:
$I = \int t \cdot \frac{2t}{t^2+4} \, dt = 2 \int \frac{t^2}{t^2+4} \, dt$.
Now,rewrite the integrand:
$I = 2 \int \frac{t^2+4-4}{t^2+4} \, dt = 2 \int \left(1 - \frac{4}{t^2+2^2}\right) \, dt$.
Integrating term by term:
$I = 2 \left[ t - 4 \cdot \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \right] + C$.
$I = 2t - 4 \tan^{-1}\left(\frac{t}{2}\right) + C$.
Substituting $t = \sqrt{e^x-4}$ back:
$I = 2 \sqrt{e^x-4} - 4 \tan^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right) + C$.

(C) Let $I = \int \frac{d x}{(x+1) \sqrt{4 x+3}}$.
Substitute $4x + 3 = t^2$,which implies $4dx = 2tdt$ or $dx = \frac{1}{2}tdt$.
Also,$x = \frac{t^2 - 3}{4}$,so $x + 1 = \frac{t^2 - 3}{4} + 1 = \frac{t^2 + 1}{4}$.
Substituting these into the integral:
$I = \int \frac{\frac{1}{2} t dt}{(\frac{t^2 + 1}{4}) t} = \int \frac{\frac{1}{2} dt}{\frac{t^2 + 1}{4}} = 2 \int \frac{dt}{t^2 + 1}$.
Integrating,we get $I = 2 \tan^{-1}(t) + c$.
Substituting back $t = \sqrt{4x + 3}$,we get $I = 2 \tan^{-1}(\sqrt{4x + 3}) + c$.

(C) Let $I = \int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x$.
Substitute $t = x^{50}$,then $dt = 50x^{49} dx$,which implies $x^{49} dx = \frac{1}{50} dt$.
Substituting these into the integral,we get:
$I = \frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} dt$.
Now,let $u = \tan ^{-1} t$,then $du = \frac{1}{1+t^2} dt$.
The integral becomes:
$I = \frac{1}{50} \int u du = \frac{1}{50} \cdot \frac{u^2}{2} + c = \frac{u^2}{100} + c$.
Substituting back $u = \tan ^{-1} (x^{50})$,we have:
$I = \frac{(\tan ^{-1} (x^{50}))^2}{100} + c$.
Comparing this with the given expression $k(\tan ^{-1} (x^{50}))^2 + c$,we find $k = \frac{1}{100}$.

(A) Let $I = \int \frac{\sin x}{\cos x(1+\cos x)} dx$.
Substitute $\cos x = t$,then $-\sin x dx = dt$,so $\sin x dx = -dt$.
$I = \int \frac{-dt}{t(1+t)} = -\int \left[ \frac{1}{t} - \frac{1}{1+t} \right] dt$.
$I = -[\log |t| - \log |1+t|] + c = \log |1+t| - \log |t| + c$.
$I = \log \left| \frac{1+t}{t} \right| + c$.
Substituting $t = \cos x$ back,we get $I = \log \left| \frac{1+\cos x}{\cos x} \right| + c$.
Since $I = f(x) + c$,we have $f(x) = \log \left| \frac{1+\cos x}{\cos x} \right|$.

(D) Let $I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sqrt{\cot x}}{\frac{\sin x \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\tan x} \sec^2 x d x$.
Since $\frac{1}{\tan x} = \cot x$,we have $I = \int \sqrt{\cot x} \cdot \cot x \cdot \sec^2 x d x = \int (\cot x)^{3/2} \sec^2 x d x$.
Alternatively,rewrite the integral as:
$I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = \int \frac{\sqrt{\cot x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\tan x \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\cot x} \cdot \frac{1}{\sin^2 x} d x$ is not correct.
Let $t = \cot x$,then $dt = -\csc^2 x d x$,so $dx = -\sin^2 x dt$.
$I = \int \frac{\sqrt{t}}{\sin x \cos x} (- \sin^2 x) dt = \int \frac{\sqrt{t}}{\cos x / \sin x} (-dt) = \int \frac{\sqrt{t}}{t} (-dt) = -\int t^{-1/2} dt$.
$I = -[2 t^{1/2}] + c = -2 \sqrt{\cot x} + c$.
Comparing with $-f(x) + c$,we get $-f(x) = -2 \sqrt{\cot x}$,so $f(x) = 2 \sqrt{\cot x}$.

(C) Let $I = \int \frac{d x}{(x+100) \sqrt{x+99}}$.
We can rewrite the denominator as $(x+99)+1$,so $I = \int \frac{d x}{((\sqrt{x+99})^2+1) \sqrt{x+99}}$.
Substitute $t = \sqrt{x+99}$,then $t^2 = x+99$,which implies $2t \, dt = dx$.
Substituting these into the integral:
$I = \int \frac{2t \, dt}{(t^2+1)t} = \int \frac{2 \, dt}{t^2+1}$.
Integrating,we get $I = 2 \tan^{-1}(t) + c$.
Substituting back $t = \sqrt{x+99}$,we get $I = 2 \tan^{-1}(\sqrt{x+99}) + c$.
Comparing this with $f(x) + c$,we find $f(x) = 2 \tan^{-1}(\sqrt{x+99})$.

(B) Let $I = \int \frac{d x}{\sqrt{x}(x+9)}$.
Substitute $x = t^2$,which implies $d x = 2t \, dt$.
Substituting these into the integral:
$I = \int \frac{2t \, dt}{t(t^2 + 9)} = \int \frac{2 \, dt}{t^2 + 3^2}$.
Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = 2 \cdot \frac{1}{3} \tan^{-1}(\frac{t}{3}) + C$.
Substituting $t = \sqrt{x}$ back into the expression:
$I = \frac{2}{3} \tan^{-1}(\frac{\sqrt{x}}{3}) + C$.

(D) Let $I = \int e^{-x} \tan ^{-1}\left(e^x\right) d x$.
Substitute $e^x = t$,then $e^x d x = d t$,which implies $d x = \frac{1}{t} d t$.
Substituting these into the integral,we get:
$I = \int \frac{\tan ^{-1} t}{t^2} d t$.
Using integration by parts,let $u = \tan ^{-1} t$ and $dv = t^{-2} dt$. Then $du = \frac{1}{1+t^2} dt$ and $v = -\frac{1}{t}$.
$I = -\frac{1}{t} \tan ^{-1} t - \int \left(-\frac{1}{t}\right) \frac{1}{1+t^2} d t = -\frac{1}{t} \tan ^{-1} t + \int \frac{1}{t(1+t^2)} d t$.
Using partial fractions,$\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.
So,$I = -\frac{1}{t} \tan ^{-1} t + \int \left(\frac{1}{t} - \frac{t}{1+t^2}\right) d t$.
$I = -\frac{1}{t} \tan ^{-1} t + \log |t| - \frac{1}{2} \log (1+t^2) + C$.
Substituting $t = e^x$ back:
$I = -e^{-x} \tan ^{-1} (e^x) + \log (e^x) - \frac{1}{2} \log (1+e^{2x}) + C$.
Since $\log (e^x) = x$,we have $I = -e^{-x} \tan ^{-1} (e^x) + x - \frac{1}{2} \log (1+e^{2x}) + C$.
Comparing this with the given expression $f(x) - \frac{1}{2} \log (1+e^{2x}) + C$,we find $f(x) = x - e^{-x} \tan ^{-1} (e^x)$.

(C) Given $f(x) = \int \frac{2 \sin x \cos x + 2 \cos x}{4 \sin^2 x + 5 \sin x + 1} \, dx$.
Factor the denominator: $4 \sin^2 x + 5 \sin x + 1 = (4 \sin x + 1)(\sin x + 1)$.
So,$f(x) = \int \frac{2 \cos x (\sin x + 1)}{(4 \sin x + 1)(\sin x + 1)} \, dx = \int \frac{2 \cos x}{4 \sin x + 1} \, dx$.
Let $u = 4 \sin x + 1$,then $du = 4 \cos x \, dx$,which implies $\cos x \, dx = \frac{du}{4}$.
$f(x) = \int \frac{2}{u} \cdot \frac{du}{4} = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |4 \sin x + 1| + C$.
Given $f(0) = 0$,we have $0 = \frac{1}{2} \ln |4 \sin(0) + 1| + C \implies 0 = \frac{1}{2} \ln(1) + C \implies C = 0$.
Thus,$f(x) = \frac{1}{2} \ln |4 \sin x + 1|$.
Now,$f\left(\frac{\pi}{6}\right) = \frac{1}{2} \ln |4 \sin(\frac{\pi}{6}) + 1| = \frac{1}{2} \ln |4(\frac{1}{2}) + 1| = \frac{1}{2} \ln(3)$.

(A) Given $f(x) = \int \frac{16x^7 + 5x^{10}}{(x^3 + 2 + 3x^8)^2} dx$.
Divide the numerator and denominator by $x^{16}$:
$f(x) = \int \frac{16x^{-9} + 5x^{-6}}{(x^{-5} + 2x^{-8} + 3)^2} dx$.
Let $u = x^{-5} + 2x^{-8} + 3$.
Then $du = (-5x^{-6} - 16x^{-9}) dx$,which implies $-du = (16x^{-9} + 5x^{-6}) dx$.
Substituting this into the integral:
$f(x) = \int -u^{-2} du = u^{-1} + C = \frac{1}{x^{-5} + 2x^{-8} + 3} + C = \frac{x^8}{1 + 2x^5 + 3x^8} + C$.
Given $f(0) = 1$,we have $1 = \frac{0}{1} + C$,so $C = 1$.
Thus,$f(x) = \frac{x^8}{1 + 2x^5 + 3x^8} + 1$.
For $f(1)$,$f(1) = \frac{1}{1 + 2 + 3} + 1 = \frac{1}{6} + 1 = \frac{7}{6}$.

(A) We have,$I = \int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x$.
Rearranging the terms in the denominator:
$x(x+3) = x^2+3x$ and $(x+1)(x+2) = x^2+3x+2$.
So,$I = \int \frac{2 x+3}{(x^2+3 x)(x^2+3 x+2)+1} d x$.
Let $t = x^2+3 x$,then $dt = (2x+3) dx$.
Substituting these into the integral:
$I = \int \frac{d t}{t(t+2)+1} = \int \frac{d t}{t^2+2 t+1} = \int \frac{d t}{(t+1)^2}$.
Integrating,we get $I = -\frac{1}{t+1} + c$.
Substituting $t = x^2+3x$ back,we get $I = -\frac{1}{x^2+3 x+1} + c$.
Comparing this with $-\frac{1}{p x^2+q x+r} + c$,we find $p=1, q=3, r=1$.
Therefore,$\frac{3 p-q}{r} = \frac{3(1)-3}{1} = \frac{0}{1} = 0$.

(B) $\int \frac{d x}{x\left(x^4+1\right)} = \int \frac{x^3 d x}{x^4\left(x^4+1\right)}$
Let $x^4 = t$,then $4x^3 dx = dt$,so $x^3 dx = \frac{dt}{4}$.
Substituting these into the integral:
$\int \frac{dt/4}{t(t+1)} = \frac{1}{4} \int \frac{dt}{t(t+1)}$
Using partial fractions:
$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$
Therefore:
$\frac{1}{4} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt = \frac{1}{4} (\log |t| - \log |t+1|) + C$
$= \frac{1}{4} \log \left| \frac{t}{t+1} \right| + C$
Substituting $t = x^4$ back:
$= \frac{1}{4} \log \left( \frac{x^4}{x^4+1} \right) + C$

(A) Let $I = \int \frac{x^3}{(1+x^2)^3} dx$.
Method $1$: Using $x = \tan \theta$,$dx = \sec^2 \theta d\theta$.
$I = \int \frac{\tan^3 \theta \sec^2 \theta}{(1+\tan^2 \theta)^3} d\theta = \int \frac{\tan^3 \theta \sec^2 \theta}{\sec^6 \theta} d\theta = \int \tan^3 \theta \cos^4 \theta d\theta = \int \sin^3 \theta \cos \theta d\theta$.
Let $\sin \theta = p$,then $\cos \theta d\theta = dp$.
$I = \int p^3 dp = \frac{p^4}{4} + k = \frac{\sin^4 \theta}{4} + k$.
Since $\tan \theta = x$,$\sin \theta = \frac{x}{\sqrt{1+x^2}}$,so $I = \frac{x^4}{4(1+x^2)^2} + k$. Thus $f(x) = \frac{x^4}{4(1+x^2)^2}$.
Method $2$: Using $1+x^2 = z$,$2x dx = dz \Rightarrow x dx = \frac{1}{2} dz$.
$I = \int \frac{x^2 \cdot x dx}{(1+x^2)^3} = \int \frac{(z-1) \cdot \frac{1}{2} dz}{z^3} = \frac{1}{2} \int (z^{-2} - z^{-3}) dz = \frac{1}{2} [-\frac{1}{z} + \frac{1}{2z^2}] + c = -\frac{1}{2(1+x^2)} + \frac{1}{4(1+x^2)^2} + c$.
Thus $g(x) = -\frac{1}{2(1+x^2)} + \frac{1}{4(1+x^2)^2}$.
Now,$f(x) - g(x) = \frac{x^4}{4(1+x^2)^2} + \frac{1}{2(1+x^2)} - \frac{1}{4(1+x^2)^2} = \frac{x^4 - 1 + 2(1+x^2)}{4(1+x^2)^2} = \frac{x^4 + 2x^2 + 1}{4(1+x^2)^2} = \frac{(x^2+1)^2}{4(1+x^2)^2} = \frac{1}{4}$.
Therefore,$f(x) - g(x) + k - c = \frac{1}{4} + k - c$,which is a constant.

(D) Given the integral: $I = \int \frac{d x}{x \ln (x) \ln ^2(x) \ldots \ln ^m(x)}$
Let $u = \ln(x)$,then $du = \frac{1}{x} dx$.
The integral becomes: $I = \int \frac{du}{u^1 \cdot u^2 \cdot u^3 \cdot \ldots \cdot u^m} = \int \frac{du}{u^{1+2+3+\ldots+m}}$.
Using the sum formula $\sum_{i=1}^{m} i = \frac{m(m+1)}{2}$,we have:
$I = \int u^{-\frac{m(m+1)}{2}} du$.
Applying the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C$:
$I = \frac{u^{-\frac{m(m+1)}{2} + 1}}{-\frac{m(m+1)}{2} + 1} + C = \frac{u^{\frac{2 - m^2 - m}{2}}}{\frac{2 - m^2 - m}{2}} + C$.
Factorizing the exponent: $2 - m^2 - m = -(m^2 + m - 2) = -(m+2)(m-1) = (m+2)(1-m)$.
So,$I = \frac{u^{\frac{(m+2)(1-m)}{2}}}{\frac{(m+2)(1-m)}{2}} + C$.
Comparing this with the given form $\frac{(\ln x)^K}{K} + C$,we identify $K = \frac{(m+2)(1-m)}{2}$.
Therefore,$2K = (m+2)(1-m)$.

(A) Given $f_n(x) = \log \log \ldots \log x$ ($n$ times).
Let $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$.
Let $t = f_n(x) = \log(f_{n-1}(x))$.
Then,$\frac{dt}{dx} = \frac{1}{f_{n-1}(x)} \cdot \frac{d}{dx}(f_{n-1}(x)) = \frac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) \cdot x}$.
Thus,$dx = (x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt$.
Substituting this into the integral:
$I = \int \frac{(x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt}{x f_1(x) f_2(x) \ldots f_{n-1}(x) f_n(x)} = \int \frac{dt}{f_n(x)} = \int \frac{dt}{t}$.
$I = \log(t) + c = \log(f_n(x)) + c$.
Since $f_{n+1}(x) = \log(f_n(x))$,we have $I = f_{n+1}(x) + c$.

(A) Let $t = \log _e(x+\sqrt{1+x^2})$.
Then,$dt = \frac{1}{x+\sqrt{1+x^2}} \cdot (1 + \frac{2x}{2\sqrt{1+x^2}}) dx = \frac{1}{x+\sqrt{1+x^2}} \cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}} dx = \frac{dx}{\sqrt{1+x^2}}$.
Substituting this into the integral,we get:
$I = \int t \, dt = \frac{t^2}{2} + c$.
Substituting $t$ back,we get $I = \frac{(\log _e(x+\sqrt{1+x^2}))^2}{2} + c$.
Comparing this with $f(g(x)) + c$,we have $f(x) = \frac{x^2}{2}$ and $g(x) = \log _e(x+\sqrt{1+x^2})$.

(B) We are given the integral $I = \int \frac{x^{1/2}}{\sqrt{1-x^3}} dx$.
Let $t = x^{3/2}$.
Then $dt = \frac{3}{2} x^{1/2} dx$,which implies $x^{1/2} dx = \frac{2}{3} dt$.
Substituting these into the integral,we get:
$I = \int \frac{\frac{2}{3} dt}{\sqrt{1-t^2}} = \frac{2}{3} \int \frac{dt}{\sqrt{1-t^2}} = \frac{2}{3} \sin^{-1}(t) + c$.
Substituting $t = x^{3/2}$ back,we get:
$I = \frac{2}{3} \sin^{-1}(x^{3/2}) + c$.
Comparing this with the given form $\frac{2}{3} g(f(x)) + c$,we identify $f(x) = x^{3/2}$ and $g(x) = \sin^{-1}(x)$.

(C) Let $I = \int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{2 \sin x \cos x}{(a+b \cos x)^{2}} d x$.
Let $t = a + b \cos x$. Then $dt = -b \sin x \, dx$,which implies $\sin x \, dx = -\frac{dt}{b}$.
Also,from $t = a + b \cos x$,we get $\cos x = \frac{t-a}{b}$.
Substituting these into the integral:
$I = \int \frac{2 (\frac{t-a}{b})}{t^2} \cdot (-\frac{dt}{b}) = -\frac{2}{b^2} \int \frac{t-a}{t^2} \, dt$.
$I = -\frac{2}{b^2} \int (\frac{1}{t} - \frac{a}{t^2}) \, dt$.
$I = -\frac{2}{b^2} [\ln |t| + \frac{a}{t}] + c$.
Substituting $t = a + b \cos x$ back:
$I = -\frac{2}{b^2} [\ln |a + b \cos x| + \frac{a}{a + b \cos x}] + c$.
Comparing this with the given expression $\alpha [\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}]+c$,we find $\alpha = -\frac{2}{b^2}$.

(C) Let $I = \int \cos (\log x) d x$.
Substitute $\log x = t$,which implies $x = e^t$.
Then,$dx = e^t dt$.
Substituting these into the integral,we get $I = \int e^t \cos t dt$.
Using the standard integration formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} [a \cos(bx) + b \sin(bx)] + C$,where $a = 1$ and $b = 1$:
$I = \frac{e^t}{1^2 + 1^2} [1 \cdot \cos t + 1 \cdot \sin t] + C$.
$I = \frac{e^t}{2} [\cos t + \sin t] + C$.
Substituting $t = \log x$ and $e^t = x$ back into the expression:
$I = \frac{x}{2} [\cos(\log x) + \sin(\log x)] + C$.
Thus,$F(x) = \frac{x}{2} [\cos(\log x) + \sin(\log x)]$.

(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} d x$.
Substitute $z = \log \sqrt{x} = \frac{1}{2} \log x$.
Then,$dz = \frac{1}{2} \cdot \frac{1}{x} dx$,which implies $\frac{dx}{x} = 2 dz$.
Substituting these into the integral:
$I = \int \frac{z}{3} (2 dz) = \frac{2}{3} \int z dz$.
Integrating $z$ with respect to $z$:
$I = \frac{2}{3} \cdot \frac{z^2}{2} + C = \frac{1}{3} z^2 + C$.
Substituting back $z = \log \sqrt{x}$:
$I = \frac{1}{3} (\log \sqrt{x})^2 + C$.

(A) Let $I = \int \frac{(x-2)}{\{(x-2)^{2}(x+3)^{7}\}^{1 / 3}} d x$.
We can rewrite the integrand as:
$I = \int \frac{(x-2)}{(x-2)^{2/3}(x+3)^{7/3}} d x = \int \frac{(x-2)^{1/3}}{(x+3)^{7/3}} d x$.
This can be written as:
$I = \int \frac{1}{(x+3)^{7/3} \cdot (x-2)^{-1/3}} d x = \int \frac{1}{(x-2)^2 \cdot \left(\frac{x+3}{x-2}\right)^{7/3}} d x$.
Let $t = \frac{x+3}{x-2}$. Then $dt = \frac{(x-2)(1) - (x+3)(1)}{(x-2)^2} dx = \frac{-5}{(x-2)^2} dx$.
Thus,$\frac{dx}{(x-2)^2} = -\frac{1}{5} dt$.
Substituting these into the integral:
$I = -\frac{1}{5} \int t^{-7/3} dt = -\frac{1}{5} \left[ \frac{t^{-4/3}}{-4/3} \right] + C$.
$I = -\frac{1}{5} \cdot \left( -\frac{3}{4} \right) t^{-4/3} + C = \frac{3}{20} t^{-4/3} + C$.
Substituting $t = \frac{x+3}{x-2}$ back:
$I = \frac{3}{20} \left( \frac{x+3}{x-2} \right)^{-4/3} + C = \frac{3}{20} \left( \frac{x-2}{x+3} \right)^{4/3} + C$.

(B) To evaluate the integral $I = \int \frac{x^3 \, dx}{1+x^8}$,we can rewrite the denominator as $1 + (x^4)^2$.
Let $u = x^4$.
Then,differentiating both sides with respect to $x$,we get $du = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{du}{4}$.
Substituting these into the integral,we get:
$I = \int \frac{du/4}{1+u^2} = \frac{1}{4} \int \frac{du}{1+u^2}$.
Using the standard integral formula $\int \frac{du}{1+u^2} = \tan^{-1}(u) + c$,we obtain:
$I = \frac{1}{4} \tan^{-1}(u) + c$.
Substituting $u = x^4$ back,we get the final result:
$I = \frac{1}{4} \tan^{-1}(x^4) + c$.

(C) Let $I = \int \frac{dx}{(e^x + e^{-x})^2}$.
Multiply the numerator and denominator by $e^{2x}$:
$I = \int \frac{e^{2x} dx}{(e^x \cdot e^x + e^{-x} \cdot e^x)^2} = \int \frac{e^{2x} dx}{(e^{2x} + 1)^2}$.
Let $u = e^{2x} + 1$. Then $du = 2e^{2x} dx$,which implies $e^{2x} dx = \frac{du}{2}$.
Substituting these into the integral:
$I = \int \frac{du/2}{u^2} = \frac{1}{2} \int u^{-2} du$.
Integrating with respect to $u$:
$I = \frac{1}{2} \left( \frac{u^{-1}}{-1} \right) + C = -\frac{1}{2u} + C$.
Substituting $u = e^{2x} + 1$ back:
$I = -\frac{1}{2(e^{2x} + 1)} + C = -\frac{1}{2}(e^{2x} + 1)^{-1} + C$.

(B) To evaluate the integral $I = \int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$,we use the method of substitution.
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 get $I = \int t dt$.
Integrating with respect to $t$,we obtain $I = \frac{1}{2} t^2 + c$.
Finally,substituting $t = \sin ^{-1} x$ back into the expression,we get $I = \frac{1}{2} (\sin ^{-1} x)^2 + c$.

(A) Let $I = \int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$.
Divide the numerator and denominator by $x^{2}$:
$I = \int \frac{1 - 1/x^{2}}{x^{2} + 3 + 1/x^{2}} d x$.
Rewrite the denominator as $(x^{2} + 1/x^{2}) + 3$:
$I = \int \frac{1 - 1/x^{2}}{(x + 1/x)^{2} - 2 + 3} d x$.
$I = \int \frac{1 - 1/x^{2}}{(x + 1/x)^{2} + 1} d x$.
Let $t = x + 1/x$. Then $dt = (1 - 1/x^{2}) d x$.
Substituting these into the integral:
$I = \int \frac{dt}{t^{2} + 1}$.
Using the standard integral formula $\int \frac{1}{t^{2} + 1} dt = \tan^{-1}(t) + C$:
$I = \tan^{-1}(t) + C$.
Substituting back $t = x + 1/x$:
$I = \tan^{-1}(x + 1/x) + C$.

(D) Let $I = \int 2^{2^{x}} \cdot 2^{x} \, dx$.
Substitute $t = 2^{x}$.
Then,$dt = 2^{x} \ln 2 \, dx$,which implies $2^{x} \, dx = \frac{dt}{\ln 2}$.
Substituting these into the integral,we get:
$I = \int 2^{t} \cdot \frac{dt}{\ln 2} = \frac{1}{\ln 2} \int 2^{t} \, dt$.
Using the formula $\int a^{t} \, dt = \frac{a^{t}}{\ln a} + C$,we have:
$I = \frac{1}{\ln 2} \cdot \frac{2^{t}}{\ln 2} + C = \frac{2^{t}}{(\ln 2)^{2}} + C$.
Substituting back $t = 2^{x}$,we get:
$I = \frac{2^{2^{x}}}{(\log 2)^{2}} + C$.
Comparing this with $A \cdot 2^{2^{x}} + C$,we find $A = \frac{1}{(\log 2)^{2}}$.

(C) Given $f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}}$.
Let $x = t^6$,then $dx = 6t^5 dt$.
Substituting these into the integral:
$f(x) = \int \frac{6t^5 dt}{t^4 + 2t^3} = \int \frac{6t^2 dt}{t + 2} = 6 \int \frac{t^2 - 4 + 4}{t + 2} dt$.
$f(x) = 6 \int (t - 2 + \frac{4}{t + 2}) dt = 6 [\frac{t^2}{2} - 2t + 4 \log_{e}(t + 2)] + C$.
Substituting $t = x^{1/6}$:
$f(x) = 3x^{1/3} - 12x^{1/6} + 24 \log_{e}(x^{1/6} + 2) + C$.
Given $f(0) = -26 + 24 \log_{e}(2)$.
At $x = 0$,$f(0) = 0 - 0 + 24 \log_{e}(2) + C = 24 \log_{e}(2) + C$.
Comparing,$C = -26$.
Now,$f(1) = 3(1)^{1/3} - 12(1)^{1/6} + 24 \log_{e}(1^{1/6} + 2) - 26$.
$f(1) = 3 - 12 + 24 \log_{e}(3) - 26 = -35 + 24 \log_{e}(3)$.
Given $f(1) = a + b \log_{e}(3)$,we have $a = -35$ and $b = 24$.
Therefore,$a + b = -35 + 24 = -11$.

(C) Given integral is $I = \int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx$.
Factor out $e^{2025}$ from the numerator and $e^{2026}$ from the denominator:
$I = \int \frac{e^{2025}(e^x - e^{-x})}{e^{2026}(e^x + e^{-x})} dx = \frac{1}{e} \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$.
Let $u = e^x + e^{-x}$.
Then,the derivative is $du = (e^x - e^{-x}) dx$.
Substituting these into the integral:
$I = \frac{1}{e} \int \frac{1}{u} du = \frac{1}{e} \ln |u| + C$.
Substituting back $u = e^x + e^{-x}$,we get:
$I = \frac{1}{e} \ln |e^x + e^{-x}| + C$.