Integration by substitution Questions in English

(B) To evaluate the integral $I = \int \frac{x}{\sqrt{x+4}} \, dx$,let $u = x+4$. Then $du = dx$ and $x = u-4$.
Substituting these into the integral,we get:
$I = \int \frac{u-4}{\sqrt{u}} \, du$
$I = \int (\frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}}) \, du$
$I = \int (u^{1/2} - 4u^{-1/2}) \, du$
Integrating term by term:
$I = \frac{u^{3/2}}{3/2} - 4 \frac{u^{1/2}}{1/2} + C$
$I = \frac{2}{3} u^{3/2} - 8 u^{1/2} + C$
$I = \frac{2}{3} u^{1/2} (u - 12) + C$
Substituting $u = x+4$ back:
$I = \frac{2}{3} \sqrt{x+4} (x+4 - 12) + C$
$I = \frac{2}{3} \sqrt{x+4} (x-8) + C$
Thus,the correct option is $B$.

(A) To evaluate the integral $I = \int \frac{\sin (\tan ^{-1} x)}{1+x^2} d x$,we use the method of substitution.
Let $u = \tan ^{-1} x$.
Then,differentiating both sides with respect to $x$,we get $du = \frac{1}{1+x^2} dx$.
Substituting these into the integral,we have $I = \int \sin(u) du$.
The integral of $\sin(u)$ is $-\cos(u) + C$.
Substituting back $u = \tan ^{-1} x$,we get $I = -\cos (\tan ^{-1} x) + C$.

(C) To solve the integral $\int \frac{1}{x + x \log x} \, dx$,we first factor the denominator:
$\int \frac{1}{x(1 + \log x)} \, dx$.
Let $u = 1 + \log x$.
Then,the derivative is $du = \frac{1}{x} \, dx$.
Substituting these into the integral,we get:
$\int \frac{1}{u} \, du = \log |u| + C$.
Substituting back $u = 1 + \log x$,we obtain:
$\log |1 + \log x| + C$.
Thus,the correct option is $C$.

(D) To evaluate the integral $I = \int \frac{dx}{e^x + e^{-x}}$,we first simplify the integrand:
$I = \int \frac{dx}{e^x + \frac{1}{e^x}} = \int \frac{e^x dx}{(e^x)^2 + 1}$.
Let $u = e^x$. Then $du = e^x dx$.
Substituting these into the integral,we get:
$I = \int \frac{du}{u^2 + 1}$.
Using the standard integral formula $\int \frac{du}{u^2 + 1} = \tan^{-1}(u) + C$,we obtain:
$I = \tan^{-1}(e^x) + C$.
Therefore,the correct option is $D$.

(C) To evaluate the integral $I = \int \frac{1}{x+x \log x} dx$,we first factor out $x$ from the denominator:
$I = \int \frac{1}{x(1+\log x)} dx$
Let $u = 1+\log x$. Then,the derivative is $du = \frac{1}{x} dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{u} du$
$I = \log |u| + C$
Substituting back $u = 1+\log x$,we obtain:
$I = \log |1+\log x| + C$
Thus,the correct option is $C$.

(A) To evaluate the integral $I = \int x^2 e^{x^3} dx$,we use the method of substitution.
Let $u = x^3$.
Then,differentiating both sides with respect to $x$,we get $du = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} du$.
Substituting these into the integral,we have:
$I = \int e^u \cdot \frac{1}{3} du$
$I = \frac{1}{3} \int e^u du$
$I = \frac{1}{3} e^u + c$
Substituting back $u = x^3$,we get:
$I = \frac{1}{3} e^{x^3} + c$.
Thus,the correct option is $A$.

(A) To solve the integral $I = \int \frac{(x^4+x)^{\frac{1}{4}}}{x^5} dx$,we factor out $x^4$ from the expression inside the parenthesis:
$I = \int \frac{(x^4(1+\frac{1}{x^3}))^{\frac{1}{4}}}{x^5} dx$
$I = \int \frac{x(1+\frac{1}{x^3})^{\frac{1}{4}}}{x^5} dx = \int \frac{(1+\frac{1}{x^3})^{\frac{1}{4}}}{x^4} dx$
Let $u = 1 + \frac{1}{x^3}$. Then $du = -\frac{3}{x^4} dx$,which implies $\frac{dx}{x^4} = -\frac{1}{3} du$.
Substituting these into the integral:
$I = \int u^{\frac{1}{4}} (-\frac{1}{3}) du = -\frac{1}{3} \int u^{\frac{1}{4}} du$
$I = -\frac{1}{3} \cdot \frac{u^{\frac{5}{4}}}{\frac{5}{4}} + C = -\frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}} + C$
$I = -\frac{4}{15} (1+\frac{1}{x^3})^{\frac{5}{4}} + C$.
Thus,the correct option is $A$.

(C) Let $I = \int \tan ^8 x \sec ^4 x \, dx$.
We can rewrite $\sec ^4 x$ as $\sec ^2 x \cdot \sec ^2 x$.
So,$I = \int \tan ^8 x \cdot \sec ^2 x \cdot \sec ^2 x \, dx$.
Since $\sec ^2 x = 1 + \tan ^2 x$,we have:
$I = \int \tan ^8 x (1 + \tan ^2 x) \sec ^2 x \, dx$.
Let $u = \tan x$,then $du = \sec ^2 x \, dx$.
Substituting these into the integral:
$I = \int u^8 (1 + u^2) \, du = \int (u^8 + u^{10}) \, du$.
Integrating with respect to $u$:
$I = \frac{u^9}{9} + \frac{u^{11}}{11} + c$.
Substituting $u = \tan x$ back:
$I = \frac{\tan ^9 x}{9} + \frac{\tan ^{11} x}{11} + c$.

(A) To solve the integral $I = \int \frac{1}{x^2 \sqrt{1-x^2}} \cdot dx$,we use the substitution $x = \sin \theta$.
Then,$dx = \cos \theta \cdot d\theta$.
Substituting these into the integral:
$I = \int \frac{\cos \theta \cdot d\theta}{\sin^2 \theta \sqrt{1-\sin^2 \theta}} = \int \frac{\cos \theta \cdot d\theta}{\sin^2 \theta \cdot \cos \theta} = \int \frac{1}{\sin^2 \theta} \cdot d\theta = \int \csc^2 \theta \cdot d\theta$.
The integral of $\csc^2 \theta$ is $-\cot \theta + C$.
Since $x = \sin \theta$,we have $\sin \theta = x$.
Then $\cos \theta = \sqrt{1-x^2}$,so $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{1-x^2}}{x}$.
Thus,$I = -\frac{\sqrt{1-x^2}}{x} + C$.
Therefore,the correct option is $A$.

(B) Let $I = \int x^2 \sqrt{8-x^6} \, dx$.
Substitute $u = x^3$,then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} du$.
The integral becomes $I = \frac{1}{3} \int \sqrt{8-u^2} \, du$.
Using the standard formula $\int \sqrt{a^2-u^2} \, du = \frac{u}{2} \sqrt{a^2-u^2} + \frac{a^2}{2} \sin^{-1} \left(\frac{u}{a}\right) + C$,where $a^2 = 8$ (so $a = 2\sqrt{2}$):
$I = \frac{1}{3} \left[ \frac{u}{2} \sqrt{8-u^2} + \frac{8}{2} \sin^{-1} \left(\frac{u}{2\sqrt{2}}\right) \right] + C$.
Substituting $u = x^3$ back:
$I = \frac{1}{3} \left[ \frac{x^3}{2} \sqrt{8-x^6} + 4 \sin^{-1} \left(\frac{x^3}{2\sqrt{2}}\right) \right] + C$.
Thus,the correct option is $B$.

(B) Let $I = \int \frac{x \cos x}{(x \sin x + \cos x)^2} dx$.
Consider $u = x \sin x + \cos x$.
Then $du = (1 \cdot \sin x + x \cos x - \sin x) dx = x \cos x dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{u^2} du = \int u^{-2} du$.
Integrating,we get $I = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C$.
Substituting back $u = x \sin x + \cos x$,we get $I = -\frac{1}{x \sin x + \cos x} + C$.
Thus,the correct option is $B$.

(B) Let $I = \int \frac{x^4+5^{x-1} \cdot \log _e 5}{x^5+5^x} \cdot d x$.
Consider the substitution $u = x^5 + 5^x$.
Then,the derivative is $du = (5x^4 + 5^x \cdot \log_e 5) \cdot dx$.
Notice that the numerator of the integrand is $x^4 + 5^{x-1} \cdot \log_e 5$.
Since $5^{x-1} = \frac{5^x}{5}$,we can rewrite the numerator as $x^4 + \frac{5^x \cdot \log_e 5}{5} = \frac{5x^4 + 5^x \cdot \log_e 5}{5} = \frac{1}{5} du$.
Substituting these into the integral,we get:
$I = \int \frac{1}{5} \cdot \frac{du}{u} = \frac{1}{5} \int \frac{du}{u} = \frac{1}{5} \log |u| + C$.
Substituting back $u = x^5 + 5^x$,we obtain $I = \frac{1}{5} \log |x^5 + 5^x| + C$.
Thus,the correct option is $B$.

(C) To evaluate the integral $ I = \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x $,we first complete the square for the quadratic expression inside the square root.
$ 3-6 x-9 x^{2} = 3 - (9 x^{2} + 6 x) = 3 - ((3 x)^{2} + 2(3 x)(1) + 1^{2} - 1) = 3 - ((3 x+1)^{2} - 1) = 4 - (3 x+1)^{2} $.
Now,the integral becomes $ I = \int \frac{1}{\sqrt{4-(3 x+1)^{2}}} d x $.
Let $ u = 3x+1 $,then $ du = 3 dx $,which implies $ dx = \frac{du}{3} $.
Substituting these into the integral,we get $ I = \int \frac{1}{\sqrt{2^{2}-u^{2}}} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{1}{\sqrt{2^{2}-u^{2}}} du $.
Using the standard formula $ \int \frac{1}{\sqrt{a^{2}-x^{2}}} dx = \sin^{-1}(\frac{x}{a}) + C $,we obtain $ I = \frac{1}{3} \sin^{-1}(\frac{u}{2}) + C $.
Substituting $ u = 3x+1 $ back,we get $ I = \frac{1}{3} \sin^{-1}(\frac{3x+1}{2}) + C $.

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.
Take $x^4$ common from the bracket: $I = \int \frac{dx}{x^2 \cdot (x^4)^{3/4} (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$.
Let $t = 1 + \frac{1}{x^4} = 1 + x^{-4}$.
Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int (1 + x^{-4})^{-3/4} (x^{-5} dx) = \int t^{-3/4} \left(-\frac{1}{4} dt\right) = -\frac{1}{4} \int t^{-3/4} dt$.
Integrating with respect to $t$:
$I = -\frac{1}{4} \left(\frac{t^{1/4}}{1/4}\right) + c = -t^{1/4} + c$.
Substituting back $t = 1 + \frac{1}{x^4} = \frac{x^4+1}{x^4}$:
$I = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + c$.

(A) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} d x$.
Substitute $u = \cos x$,then $du = -\sin x d x$,or $\sin x d x = -du$.
Substituting these into the integral gives $I = \int \frac{-du}{3+4u^2} = -\int \frac{du}{3+(2u)^2}$.
Factor out $3$ from the denominator: $I = -\frac{1}{3} \int \frac{du}{1 + (\frac{2u}{\sqrt{3}})^2}$.
Let $t = \frac{2u}{\sqrt{3}}$,then $dt = \frac{2}{\sqrt{3}} du$,so $du = \frac{\sqrt{3}}{2} dt$.
Substituting $t$ into the integral: $I = -\frac{1}{3} \int \frac{\frac{\sqrt{3}}{2} dt}{1+t^2} = -\frac{\sqrt{3}}{6} \int \frac{dt}{1+t^2}$.
Since $\frac{\sqrt{3}}{6} = \frac{1}{2\sqrt{3}}$,we have $I = -\frac{1}{2\sqrt{3}} \tan^{-1}(t) + C$.
Substituting back $t = \frac{2 \cos x}{\sqrt{3}}$,we get $I = -\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right) + C$.

(A) Let $I = \int \sqrt{\operatorname{cosec} x - \sin x} \, dx$
$I = \int \sqrt{\frac{1}{\sin x} - \sin x} \, dx = \int \sqrt{\frac{1 - \sin^2 x}{\sin x}} \, dx$
$I = \int \frac{\sqrt{\cos^2 x}}{\sqrt{\sin x}} \, dx = \int \frac{\cos x}{\sqrt{\sin x}} \, dx$
Let $u = \sin x$,then $du = \cos x \, dx$
$I = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} \, du$
$I = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$
Substituting $u = \sin x$,we get $I = 2\sqrt{\sin x} + C$

(B) Let $I = \int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x$.
Dividing the numerator and denominator by $\cos ^2 x$,we get:
$I = \int \frac{\sec ^2 x}{\sec ^2 x+3 \tan ^2 x+8} d x$.
Since $\sec ^2 x = 1 + \tan ^2 x$,we have:
$I = \int \frac{\sec ^2 x}{1 + \tan ^2 x + 3 \tan ^2 x + 8} d x = \int \frac{\sec ^2 x}{4 \tan ^2 x + 9} d x$.
Let $\tan x = t$,then $\sec ^2 x d x = d t$.
Substituting these into the integral:
$I = \int \frac{d t}{4 t^2 + 9} = \frac{1}{4} \int \frac{d t}{t^2 + (3/2)^2}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{1}{4} \times \frac{1}{3/2} \tan^{-1}(\frac{t}{3/2}) + C = \frac{1}{4} \times \frac{2}{3} \tan^{-1}(\frac{2t}{3}) + C$.
$I = \frac{1}{6} \tan^{-1}(\frac{2 \tan x}{3}) + C$.

(A) Let $I = \int \frac{x^{3} \sin \left(\tan ^{-1}\left(x^{4}\right)\right)}{1+x^{8}} dx$.
Substitute $t = \tan ^{-1}\left(x^{4}\right)$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{1+(x^{4})^{2}} \cdot 4x^{3} = \frac{4x^{3}}{1+x^{8}}$.
Thus,$\frac{x^{3}}{1+x^{8}} dx = \frac{1}{4} dt$.
Substituting these into the integral,we get $I = \int \sin(t) \cdot \frac{1}{4} dt$.
$I = \frac{1}{4} \int \sin(t) dt = \frac{1}{4} (-\cos(t)) + C$.
Substituting back $t = \tan ^{-1}\left(x^{4}\right)$,we get $I = -\frac{1}{4} \cos \left(\tan ^{-1}\left(x^{4}\right)\right) + C$.

(C) Let $I = \int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} dx$.
Substitute $x^{3} = t$.
Then,$3x^{2} dx = dt$,which implies $x^{2} dx = \frac{1}{3} dt$.
Substituting these into the integral,we get:
$I = \frac{1}{3} \int \frac{1}{\sqrt{t^{2} + (a^{3})^{2}}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^{2} + a^{2}}} dx = \log |x + \sqrt{x^{2} + a^{2}}| + C$,we have:
$I = \frac{1}{3} \log |t + \sqrt{t^{2} + (a^{3})^{2}}| + C$.
Substituting $t = x^{3}$ back into the expression:
$I = \frac{1}{3} \log |x^{3} + \sqrt{x^{6} + a^{6}}| + C$.

(B) Let $ I = \int \frac{1}{\sqrt{x} + x \sqrt{x}} dx $.
Factor out $ \sqrt{x} $ from the denominator: $ I = \int \frac{1}{\sqrt{x}(1 + x)} dx $.
Since $ x = (\sqrt{x})^2 $,we have $ I = \int \frac{1}{\sqrt{x}(1 + (\sqrt{x})^2)} dx $.
Let $ t = \sqrt{x} $. Then $ dt = \frac{1}{2\sqrt{x}} dx $,which implies $ \frac{1}{\sqrt{x}} dx = 2 dt $.
Substituting these into the integral: $ I = \int \frac{2}{1 + t^2} dt $.
Integrating,we get $ I = 2 \tan^{-1}(t) + C $.
Substituting $ t = \sqrt{x} $ back,we get $ I = 2 \tan^{-1}(\sqrt{x}) + C $.

(B) Let $I = \int \frac{e^{x}(1+x) dx}{\cos^{2}(x e^{x})}$.
Substitute $t = x e^{x}$.
Then,$dt = (1 \cdot e^{x} + x \cdot e^{x}) dx = e^{x}(1+x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{\cos^{2} t}$.
$I = \int \sec^{2} t dt$.
Integrating $\sec^{2} t$,we get $I = \tan t + c$.
Substituting back $t = x e^{x}$,we get $I = \tan(x e^{x}) + c$.

(C) Let $I = \int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} d x$.
We can rewrite the integrand as:
$I = \int \frac{\cos ^{n-1} x}{\sin ^{n-1} x \cdot \sin^2 x} d x = \int \cot^{n-1} x \cdot \csc^2 x d x$.
Let $t = \cot x$.
Then $dt = -\csc^2 x d x$,which implies $\csc^2 x d x = -dt$.
Substituting these into the integral:
$I = \int t^{n-1} (-dt) = -\int t^{n-1} dt$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
$I = -\frac{t^n}{n} + C$.
Substituting $t = \cot x$ back:
$I = -\frac{\cot^n x}{n} + C$.

(D) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$.
Substitute $x = \tan^2 \theta$,which implies $dx = 2 \tan \theta \sec^2 \theta d\theta$.
Then,$\sqrt{x} = \tan \theta$.
Substituting these into the integral:
$I = \int \frac{\tan \theta}{\tan^2 \theta (\tan^2 \theta + 1)} \cdot (2 \tan \theta \sec^2 \theta) d\theta$.
Since $\tan^2 \theta + 1 = \sec^2 \theta$,the expression simplifies to:
$I = \int \frac{\tan \theta}{\tan^2 \theta \sec^2 \theta} \cdot (2 \tan \theta \sec^2 \theta) d\theta$.
$I = \int \frac{2 \tan^2 \theta \sec^2 \theta}{\tan^2 \theta \sec^2 \theta} d\theta = \int 2 d\theta$.
$I = 2\theta + C$.
Since $x = \tan^2 \theta$,we have $\theta = \tan^{-1}(\sqrt{x})$.
Therefore,$I = 2 \tan^{-1}(\sqrt{x}) + C$.
Comparing this with $k \tan^{-1} m$,we get $k = 2$ and $m = \sqrt{x}$.
Thus,$(k, m) = (2, \sqrt{x})$.

(A) Let $I = \int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} dx$.
Substitute $\sin^{2} x = t$.
Differentiating both sides with respect to $x$,we get $2 \sin x \cos x dx = dt$,which implies $\sin x \cos x dx = \frac{1}{2} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{2 \sqrt{1-t^{2}}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{1-t^{2}}} dt = \sin^{-1} t + C$,we obtain $I = \frac{1}{2} \sin^{-1} t + C$.
Finally,substituting $t = \sin^{2} x$ back,we get $I = \frac{1}{2} \sin^{-1}(\sin^{2} x) + C$.

(B) We have,$I = \int \frac{1+x^{4}}{1+x^{6}} dx$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{\frac{1}{x^2} + x^2}{\frac{1}{x^2} + x^4} dx$.
Alternatively,we can write:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}}{1+x^{6}} dx - \int \frac{x^{2}}{1+x^{6}} dx$.
However,a simpler approach is:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx$ is not direct.
Let's use the substitution method:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx$.
Actually,the standard way is:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx$.
Using the identity $1+x^6 = (1+x^2)(1-x^2+x^4)$:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{(1+x^{2})(1-x^{2}+x^{4})} dx = \int \frac{1}{1+x^2} dx + \int \frac{x^2(x^2-1)}{1+x^6} dx$.
Following the provided steps:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx$ is complex.
Correct approach:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx = \tan^{-1} x + \int \frac{x^2}{1+(x^3)^2} dx$.
Let $t = x^3$,then $dt = 3x^2 dx$,so $x^2 dx = \frac{dt}{3}$.
Thus,the integral becomes $\tan^{-1} x + \frac{1}{3} \int \frac{dt}{1+t^2} = \tan^{-1} x + \frac{1}{3} \tan^{-1} t + C = \tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + C$.

(A) Let $I = \int \frac{3^x}{\sqrt{1-9^x}} d x$.
We can rewrite the denominator as $9^x = (3^x)^2$,so $I = \int \frac{3^x}{\sqrt{1-(3^x)^2}} d x$.
Substitute $t = 3^x$. Then,differentiating with respect to $x$,we get $dt = 3^x \log 3 \, dx$,which implies $3^x \, dx = \frac{1}{\log 3} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{\sqrt{1-t^2}} \cdot \frac{1}{\log 3} dt$.
Since $\int \frac{1}{\sqrt{1-t^2}} dt = \sin^{-1}(t) + c$,we have $I = \frac{1}{\log 3} \sin^{-1}(t) + c$.
Finally,substituting $t = 3^x$ back,we get $I = \frac{1}{\log 3} \sin^{-1}(3^x) + c$ or $\sin^{-1}(3^x) \cdot (\log 3)^{-1} + c$.

(B) We are given the integral $I = \int \frac{\sin ^6 x}{\cos ^8 x} d x$.
Rewrite the integrand as:
$I = \int \frac{\sin ^6 x}{\cos ^6 x} \cdot \frac{1}{\cos ^2 x} d x$.
Since $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos x} = \sec x$,we have:
$I = \int \tan ^6 x \sec ^2 x d x$.
Let $u = \tan x$,then $du = \sec ^2 x d x$.
Substituting these into the integral,we get:
$I = \int u ^6 d u = \frac{u ^7}{7} + c$.
Substituting back $u = \tan x$,we obtain:
$I = \frac{\tan ^7 x}{7} + c$.

(C) Given $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 2x^7 + 1)^2} dx$.
Divide the numerator and denominator by $x^{14}$:
$f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + 2 + x^{-7})^2} dx$.
Let $u = x^{-5} + x^{-7} + 2$.
Then $du = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = du$.
Substituting this into the integral:
$f(x) = -\int \frac{1}{u^2} du = \frac{1}{u} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C$.
Since $f(0) = 0$,we evaluate the limit as $x \to 0^+$.
As $x \to 0^+$,$x^{-5} + x^{-7} \to \infty$,so $\frac{1}{x^{-5} + x^{-7} + 2} \to 0$.
Thus,$0 = 0 + C$,which gives $C = 0$.
Therefore,$f(x) = \frac{1}{x^{-5} + x^{-7} + 2} = \frac{x^7}{1 + x^2 + 2x^7}$.
Evaluating at $x = 1$:
$f(1) = \frac{1^7}{1 + 1^2 + 2(1)^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.

(A) Given that $f(x)$ is an anti-derivative of $g(x)$,we have $f'(x) = g(x)$.
We need to evaluate the integral $F(x) = \int f(x) g(x) (1 + f^2(x)) dx$.
Let $u = 1 + f^2(x)$.
Then,differentiating with respect to $x$,we get $du = 2 f(x) f'(x) dx$.
Since $f'(x) = g(x)$,we have $du = 2 f(x) g(x) dx$,which implies $f(x) g(x) dx = \frac{du}{2}$.
Substituting these into the integral,we get $F(x) = \int u \cdot \frac{du}{2} = \frac{1}{2} \int u du$.
Integrating,we get $F(x) = \frac{1}{2} \cdot \frac{u^2}{2} + C = \frac{u^2}{4} + C$.
Substituting back $u = 1 + f^2(x)$,we get $F(x) = \frac{(1 + f^2(x))^2}{4} + C$.

(A) Let $I = \int \frac{d x}{(1+\sqrt{x})^{2022}}$.
Substitute $x = t^2$,then $dx = 2t dt$.
The integral becomes $I = \int \frac{2t dt}{(1+t)^{2022}}$.
Rewrite the numerator as $t = (t+1) - 1$:
$I = 2 \int \frac{t+1-1}{(1+t)^{2022}} dt = 2 \left[ \int \frac{t+1}{(1+t)^{2022}} dt - \int \frac{1}{(1+t)^{2022}} dt \right]$.
$I = 2 \left[ \int \frac{1}{(1+t)^{2021}} dt - \int \frac{1}{(1+t)^{2022}} dt \right]$.
Integrating both terms:
$I = 2 \left[ \frac{(1+t)^{-2020}}{-2020} - \frac{(1+t)^{-2021}}{-2021} \right] + C$.
$I = 2 \left[ \frac{-1}{2020(1+t)^{2020}} + \frac{1}{2021(1+t)^{2021}} \right] + C$.
Factor out $\frac{1}{(1+t)^{2021}}$:
$I = \frac{2}{(1+t)^{2021}} \left[ \frac{-(1+t)}{2020} + \frac{1}{2021} \right] + C$.
Substituting $t = \sqrt{x}$ back:
$I = \frac{2}{(1+\sqrt{x})^{2021}} \left[ \frac{-(1+\sqrt{x})}{2020} + \frac{1}{2021} \right] + C$.

(D) Given integral,$I = \int \frac{d x}{x+\sqrt{x-1}}$
Put $x-1 = t^2 \Rightarrow dx = 2t \, dt$
Then $I = \int \frac{2t}{(t^2+1)+t} \, dt = \int \frac{(2t+1)-1}{t^2+t+1} \, dt$
$= \int \frac{2t+1}{t^2+t+1} \, dt - \int \frac{dt}{t^2+t+1}$
$= \log _e|t^2+t+1| - \int \frac{dt}{(t+\frac{1}{2})^2 + \frac{3}{4}}$
$= \log _e|t^2+t+1| - \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + c$
$= \log _e|t^2+t+1| - \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2t+1}{\sqrt{3}}\right) + c$
On substituting $t = \sqrt{x-1}$,we get
$I = \log _e|x+\sqrt{x-1}| - \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}\right) + c$
Hence,option $D$ is correct.

(C) Let $I = \int \frac{\sin 2x \, dx}{\sin^4 x + \cos^4 x}$.
We know that $\sin 2x = 2 \sin x \cos x$,so $I = \int \frac{2 \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$.
Divide the numerator and denominator by $\cos^4 x$:
$I = \int \frac{2 \sin x \cos x / \cos^4 x}{(\sin^4 x + \cos^4 x) / \cos^4 x} \, dx = \int \frac{2 \tan x \sec^2 x}{1 + \tan^4 x} \, dx$.
Let $t = \tan^2 x$,then $dt = 2 \tan x \sec^2 x \, dx$.
Substituting these into the integral:
$I = \int \frac{dt}{1 + t^2} = \tan^{-1}(t) + c = \tan^{-1}(\tan^2 x) + c$.
Comparing this with $\tan^{-1}(f(x)) + c$,we get $f(x) = \tan^2 x$.
Therefore,$f\left(\frac{\pi}{3}\right) = \tan^2\left(\frac{\pi}{3}\right) = (\sqrt{3})^2 = 3$.

(C) Let $I = \int \frac{x}{(1-x^2) \sqrt{2-x^2}} dx$.
Substitute $t = \sqrt{2-x^2}$.
Then $t^2 = 2-x^2$,which implies $x^2 = 2-t^2$.
Differentiating both sides,$2t dt = -2x dx$,so $x dx = -t dt$.
Substituting these into the integral:
$I = \int \frac{-t dt}{(1-(2-t^2)) t} = \int \frac{-dt}{t^2-1} = \int \frac{dt}{1-t^2}$.
Using the standard integral formula $\int \frac{dt}{a^2-t^2} = \frac{1}{2a} \log \left| \frac{a+t}{a-t} \right| + c$ with $a=1$:
$I = \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + c$.
Substituting $t = \sqrt{2-x^2}$ back:
$I = \frac{1}{2} \log \left| \frac{1+\sqrt{2-x^2}}{1-\sqrt{2-x^2}} \right| + c$.
Thus,the correct option is $C$.

(D) Let $I = \int \frac{\sin x+\cos x}{\sin x-\cos x} d x$.
Consider the substitution $u = \sin x - \cos x$.
Then,$du = (\cos x - (-\sin x)) dx = (\cos x + \sin x) dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{u} du$.
Integrating with respect to $u$,we have:
$I = \log |u| + c$.
Substituting back $u = \sin x - \cos x$,we get:
$I = \log |\sin x - \cos x| + c$.

(D) Let $I = \int \frac{(3 x-2) \tan \left(\sqrt{9 x^2-12 x+1}\right)}{\sqrt{9 x^2-12 x+1}} d x$.
Let $u = \sqrt{9 x^2-12 x+1}$.
Then $u^2 = 9 x^2-12 x+1$.
Differentiating both sides with respect to $x$,we get $2u \frac{du}{dx} = 18x - 12 = 6(3x - 2)$.
Thus,$u \frac{du}{dx} = 3(3x - 2)$,which implies $(3x - 2) dx = \frac{1}{3} u du$.
Substituting these into the integral,we get $I = \int \frac{\tan(u)}{u} \cdot \frac{1}{3} u du = \frac{1}{3} \int \tan(u) du$.
The integral of $\tan(u)$ is $\log |\sec(u)| + c$.
Therefore,$I = \frac{1}{3} \log |\sec(\sqrt{9 x^2-12 x+1})| + c$.

(C) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin ^4 x} d x$.
Substitute $u = \sin x$,then $du = \cos x dx$.
The integral becomes $I = \int \frac{\cos ^2 x}{\sin ^2 x+\sin ^4 x} \cos x dx = \int \frac{1-u^2}{u^2+u^4} du$.
$I = \int \frac{1-u^2}{u^2(1+u^2)} du$.
Using partial fractions,$\frac{1-u^2}{u^2(1+u^2)} = \frac{A}{u} + \frac{B}{u^2} + \frac{Cu+D}{1+u^2}$.
Solving for constants,we get $\frac{1-u^2}{u^2(1+u^2)} = \frac{1}{u^2} - \frac{2}{1+u^2}$.
Thus,$I = \int (u^{-2} - 2(1+u^2)^{-1}) du = -u^{-1} - 2 \tan^{-1}(u) + c$.
Substituting back $u = \sin x$,$I = -\frac{1}{\sin x} - 2 \tan^{-1}(\sin x) + c = c - \operatorname{cosec} x - 2 \tan^{-1}(\sin x)$.
Comparing with the given form $c - \operatorname{cosec} x - f(x)$,we have $f(x) = 2 \tan^{-1}(\sin x)$.
Therefore,$f\left(\frac{\pi}{2}\right) = 2 \tan^{-1}(\sin \frac{\pi}{2}) = 2 \tan^{-1}(1) = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$.

(NONE) Let $I = \int \frac{\sec^2 x}{(\sec x + \tan x)^{5/2}} dx$.
We know that $\sec x + \tan x = t$.
Differentiating with respect to $x$,we get $(\sec x \tan x + \sec^2 x) dx = dt$,which implies $\sec x(\tan x + \sec x) dx = dt$.
Thus,$\sec x dx = \frac{dt}{t}$.
Also,$\sec x - \tan x = \frac{1}{t}$.
Adding the two equations: $2 \sec x = t + \frac{1}{t} = \frac{t^2+1}{t}$,so $\sec x = \frac{t^2+1}{2t}$.
Substituting these into the integral:
$I = \int \frac{\sec x \cdot \sec x dx}{t^{5/2}} = \int \frac{(\frac{t^2+1}{2t}) \cdot \frac{dt}{t}}{t^{5/2}} = \frac{1}{2} \int \frac{t^2+1}{t^{7/2}} dt$.
$I = \frac{1}{2} \int (t^{-3/2} + t^{-7/2}) dt$.
Integrating,we get $I = \frac{1}{2} [\frac{t^{-1/2}}{-1/2} + \frac{t^{-5/2}}{-5/2}] + c = -t^{-1/2} - \frac{1}{5} t^{-5/2} + c$.
Substituting $t = \sec x + \tan x$,we get $I = -(\sec x + \tan x)^{-1/2} - \frac{1}{5}(\sec x + \tan x)^{-5/2} + c$.

(C) Let $I = \int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx = \int \left( \sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} \right) \, dx = \int \frac{\tan x + 1}{\sqrt{\tan x}} \, dx$.
Substitute $\sqrt{\tan x} = t$,so $\tan x = t^2$ and $\sec^2 x \, dx = 2t \, dt$.
Since $\sec^2 x = 1 + \tan^2 x = 1 + t^4$,we have $dx = \frac{2t}{1 + t^4} \, dt$.
Substituting these into the integral: $I = \int \frac{t^2 + 1}{t} \cdot \frac{2t}{1 + t^4} \, dt = 2 \int \frac{t^2 + 1}{t^4 + 1} \, dt$.
Divide numerator and denominator by $t^2$: $I = 2 \int \frac{1 + 1/t^2}{t^2 + 1/t^2} \, dt = 2 \int \frac{1 + 1/t^2}{(t - 1/t)^2 + 2} \, dt$.
Let $u = t - 1/t$,then $du = (1 + 1/t^2) \, dt$.
$I = 2 \int \frac{du}{u^2 + (\sqrt{2})^2} = 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{u}{\sqrt{2}} \right) + c = \sqrt{2} \tan^{-1} \left( \frac{t - 1/t}{\sqrt{2}} \right) + c$.
Substituting $t = \sqrt{\tan x}$: $I = \sqrt{2} \tan^{-1} \left( \frac{\tan x - 1}{\sqrt{2 \tan x}} \right) + c$.

(B) Let $I = \int \frac{\sqrt{x-2}}{2x+4} dx$.
Substitute $t = \sqrt{x-2}$,then $t^2 = x-2$,so $x = t^2+2$ and $dx = 2t dt$.
Substituting these into the integral:
$I = \int \frac{t}{2(t^2+2)+4} (2t dt) = \int \frac{2t^2}{2t^2+8} dt = \int \frac{t^2}{t^2+4} dt$.
Now,rewrite the integrand:
$I = \int \frac{t^2+4-4}{t^2+4} dt = \int \left(1 - \frac{4}{t^2+4}\right) dt$.
Integrating term by term:
$I = \int 1 dt - 4 \int \frac{1}{t^2+2^2} dt = t - 4 \left(\frac{1}{2} \operatorname{Tan}^{-1}\left(\frac{t}{2}\right)\right) + c$.
$I = t - 2 \operatorname{Tan}^{-1}\left(\frac{t}{2}\right) + c$.
Substituting $t = \sqrt{x-2}$ back:
$I = \sqrt{x-2} - 2 \operatorname{Tan}^{-1}\left(\frac{\sqrt{x-2}}{2}\right) + c$.

(C) Let $I = \int \frac{x+1}{(x-2) \sqrt{1-x}} d x$.
Put $1-x = t^2$,so $x = 1-t^2$ and $dx = -2t dt$.
Substituting these into the integral:
$I = \int \frac{(1-t^2)+1}{(1-t^2-2) \cdot t} (-2t) dt$
$I = \int \frac{2-t^2}{(-1-t^2) \cdot t} (-2t) dt$
$I = \int \frac{2-t^2}{-(1+t^2)} (-2) dt = 2 \int \frac{2-t^2}{1+t^2} dt$
$I = 2 \int \frac{3-(1+t^2)}{1+t^2} dt = 2 \int \left( \frac{3}{1+t^2} - 1 \right) dt$
$I = 6 \int \frac{1}{1+t^2} dt - 2 \int 1 dt$
$I = 6 \operatorname{Tan}^{-1}(t) - 2t + c$
Substituting $t = \sqrt{1-x}$ back:
$I = 6 \operatorname{Tan}^{-1}(\sqrt{1-x}) - 2\sqrt{1-x} + c$.

(B) Let $I = \int \frac{\operatorname{cosec} x}{3 \cos x + 4 \sin x} dx$.
Divide the numerator and denominator by $\sin x$:
$I = \int \frac{\operatorname{cosec}^2 x}{3 \cot x + 4} dx$.
Let $t = 3 \cot x + 4$.
Then $dt = -3 \operatorname{cosec}^2 x dx$,which implies $\operatorname{cosec}^2 x dx = -\frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \frac{-1/3}{t} dt = -\frac{1}{3} \ln |t| + C$.
Substituting back $t = 3 \cot x + 4$:
$I = -\frac{1}{3} \ln |3 \cot x + 4| + C = -\frac{1}{3} \ln \left| \frac{3 \cos x + 4 \sin x}{\sin x} \right| + C$.
Using the property $-\ln |x| = \ln |1/x|$:
$I = \frac{1}{3} \ln \left| \frac{\sin x}{3 \cos x + 4 \sin x} \right| + C$.

(B) Let $I = \int \frac{3 x^9+7 x^8}{(x^2+2 x+5 x^8)^2} dx$.
Divide the numerator and denominator by $x^{16}$ inside the square:
$I = \int \frac{3 x^9+7 x^8}{x^{16} (x^{-6} + 2x^{-7} + 5)^2} dx = \int \frac{3 x^{-7} + 7 x^{-8}}{(x^{-6} + 2x^{-7} + 5)^2} dx$.
Let $t = x^{-6} + 2x^{-7} + 5$.
Then $dt = (-6x^{-7} - 14x^{-8}) dx = -2(3x^{-7} + 7x^{-8}) dx$.
So,$(3x^{-7} + 7x^{-8}) dx = -\frac{1}{2} dt$.
Substituting this into the integral:
$I = \int \frac{-1/2}{t^2} dt = -\frac{1}{2} \int t^{-2} dt = -\frac{1}{2} (\frac{t^{-1}}{-1}) + C = \frac{1}{2t} + C$.
Substituting $t$ back:
$I = \frac{1}{2(x^{-6} + 2x^{-7} + 5)} + C = \frac{1}{2(\frac{x+2+5x^7}{x^7})} + C = \frac{x^7}{2(5x^7+x+2)} + C$.

(D) Given the integral $I = \int \frac{3}{2 \cos ^3 x \sqrt{2 \sin 2 x}} d x$.
Using $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$,we have $\sqrt{2 \sin 2x} = \sqrt{\frac{4 \tan x}{1 + \tan^2 x}} = \frac{2 \sqrt{\tan x}}{\sec x}$.
Substituting this into the integral:
$I = \int \frac{3}{2 \cos^3 x \cdot \frac{2 \sqrt{\tan x}}{\sec x}} d x = \int \frac{3 \sec^2 x}{4 \sqrt{\tan x}} d x$.
Since $\sec^2 x = 1 + \tan^2 x$,we rewrite the integral as:
$I = \int \frac{3(1 + \tan^2 x)}{4 \sqrt{\tan x}} \sec^2 x d x$.
Let $t = \tan x$,then $dt = \sec^2 x d x$.
$I = \frac{3}{4} \int (t^{-1/2} + t^{3/2}) dt = \frac{3}{4} [2t^{1/2} + \frac{2}{5} t^{5/2}] + c = \frac{3}{2} (\tan x)^{1/2} + \frac{3}{10} (\tan x)^{5/2} + c$.
Comparing this with the given form $\frac{3}{2}(\tan x)^B + \frac{1}{10}(\tan x)^A + c$,we note that the coefficient of $(\tan x)^{5/2}$ is $\frac{3}{10}$.
Wait,the problem states $\frac{1}{10}(\tan x)^A$. If the result is $\frac{3}{10}(\tan x)^{5/2}$,there might be a typo in the question's coefficient. Assuming the form is $\frac{3}{10}(\tan x)^A$,then $A = \frac{5}{2}$.

(D) To solve the integral $I = \int \frac{dx}{x(x^4+1)}$,multiply the numerator and denominator by $x^3$:
$I = \int \frac{x^3 dx}{x^4(x^4+1)}$
Let $x^4 = t$,then $4x^3 dx = dt$,which implies $x^3 dx = \frac{dt}{4}$.
Substituting these into the integral:
$I = \frac{1}{4} \int \frac{dt}{t(t+1)}$
Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
$I = \frac{1}{4} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$
$I = \frac{1}{4} [\log|t| - \log|t+1|] + C$
$I = \frac{1}{4} \log \left| \frac{t}{t+1} \right| + C$
Substituting $t = x^4$ back:
$I = \frac{1}{4} \log \left( \frac{x^4}{x^4+1} \right) + C$

(D) $I = \int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}$
Using the identity $\cos(x-\alpha) = \cos x \cos \alpha + \sin x \sin \alpha$,we get:
$I = \int \frac{d x}{\sqrt{\sin ^3 x (\cos x \cos \alpha + \sin x \sin \alpha)}}$
Taking $\sin x$ common from the bracket:
$I = \int \frac{d x}{\sqrt{\sin ^4 x (\cot x \cos \alpha + \sin \alpha)}} = \int \frac{\csc^2 x}{\sqrt{\cot x \cos \alpha + \sin \alpha}} d x$
Let $t = \cot x \cos \alpha + \sin \alpha$.
Then $dt = -\csc^2 x \cos \alpha \, dx$,which implies $\csc^2 x \, dx = -\frac{dt}{\cos \alpha}$.
Substituting these into the integral:
$I = \int \frac{-dt}{\cos \alpha \sqrt{t}} = -\frac{1}{\cos \alpha} \int t^{-1/2} dt$
$I = -\frac{1}{\cos \alpha} (2 \sqrt{t}) + C = -\frac{2}{\cos \alpha} \sqrt{\cot x \cos \alpha + \sin \alpha} + C$
Factoring out $\cos \alpha$ from the square root:
$I = -\frac{2}{\cos \alpha} \sqrt{\cos \alpha (\cot x + \frac{\sin \alpha}{\cos \alpha})} + C$
$I = -\frac{2}{\sqrt{\cos \alpha}} \sqrt{\cot x + \tan \alpha} + C$

(C) Let $I = \int \frac{e^{2x}}{(e^x+1)^{1/4}} dx$.
Substitute $t = (e^x+1)^{1/4}$,then $t^4 = e^x+1$,so $e^x = t^4-1$.
Differentiating both sides,$e^x dx = 4t^3 dt$.
Substituting these into the integral:
$I = \int \frac{(t^4-1) \cdot (4t^3 dt)}{t} = 4 \int (t^4-1)t^2 dt$.
$I = 4 \int (t^6-t^2) dt = 4 \left( \frac{t^7}{7} - \frac{t^3}{3} \right) + C$.
$I = 4t^3 \left( \frac{t^4}{7} - \frac{1}{3} \right) + C$.
Substitute $t = (e^x+1)^{1/4}$ back:
$I = 4(e^x+1)^{3/4} \left( \frac{e^x+1}{7} - \frac{1}{3} \right) + C$.
$I = 4(e^x+1)^{3/4} \left( \frac{3e^x+3-7}{21} \right) + C$.
$I = \frac{4}{21}(e^x+1)^{3/4}(3e^x-4) + C$.

(A) $I = \int \sin ^{-1} \sqrt{\frac{x}{a+x}} dx$
Let $x = a \tan^2 t$,then $dx = 2a \tan t \sec^2 t dt$.
Substituting these into the integral:
$I = \int \sin^{-1} \sqrt{\frac{a \tan^2 t}{a(1 + \tan^2 t)}} (2a \tan t \sec^2 t) dt$
$I = \int \sin^{-1} \sqrt{\frac{\tan^2 t}{\sec^2 t}} (2a \tan t \sec^2 t) dt$
$I = \int t (2a \tan t \sec^2 t) dt = 2a \int t \tan t \sec^2 t dt$
Using integration by parts:
$I = 2a \left[ t \int \tan t \sec^2 t dt - \int (1 \cdot \int \tan t \sec^2 t dt) dt \right]$
Since $\int \tan t \sec^2 t dt = \frac{\tan^2 t}{2}$:
$I = 2a \left[ t \cdot \frac{\tan^2 t}{2} - \int \frac{\tan^2 t}{2} dt \right] = a t \tan^2 t - a \int (\sec^2 t - 1) dt$
$I = a t \tan^2 t - a (\tan t - t) + C = a t \tan^2 t - a \tan t + at + C$
$I = a t (\tan^2 t + 1) - a \tan t + C = a t \sec^2 t - a \tan t + C$
Substituting $t = \tan^{-1} \sqrt{\frac{x}{a}}$ and $\tan^2 t = \frac{x}{a}$:
$I = a \tan^{-1} \sqrt{\frac{x}{a}} (1 + \frac{x}{a}) - a \sqrt{\frac{x}{a}} + C$
$I = (a+x) \tan^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + C$

(D) Let $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$.
Divide the numerator and the denominator by $x^5$:
$I = \int \frac{\frac{1}{x^3} - \frac{1}{x^5}}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} d x$.
Let $t = 2 - \frac{2}{x^2} + \frac{1}{x^4}$.
Then $dt = (\frac{4}{x^3} - \frac{4}{x^5}) d x$,which implies $(\frac{1}{x^3} - \frac{1}{x^5}) d x = \frac{dt}{4}$.
Substituting these into the integral:
$I = \frac{1}{4} \int \frac{1}{\sqrt{t}} dt = \frac{1}{4} (2\sqrt{t}) + c = \frac{\sqrt{t}}{2} + c$.
Substituting $t = 2 - \frac{2}{x^2} + \frac{1}{x^4} = \frac{2x^4 - 2x^2 + 1}{x^4}$ back:
$I = \frac{1}{2} \sqrt{\frac{2x^4 - 2x^2 + 1}{x^4}} + c = \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2} + c$.

(A) Let $I = \int \frac{x^3 \tan^{-1} x^4}{1+x^8} dx$.
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 $I = \frac{1}{4} \int \frac{\tan^{-1} t}{1+t^2} dt$.
Now,let $u = \tan^{-1} t$,then $du = \frac{1}{1+t^2} dt$.
Substituting $u$ into the integral,we get $I = \frac{1}{4} \int u du$.
Integrating with respect to $u$,we get $I = \frac{1}{4} \cdot \frac{u^2}{2} + c = \frac{u^2}{8} + c$.
Substituting back $u = \tan^{-1} t$ and $t = x^4$,we get $I = \frac{(\tan^{-1}(x^4))^2}{8} + c$.

(A) Let $I = \int \frac{1}{x^2\sqrt{1+x^2}} dx$.
Take $x^2$ common from the square root: $I = \int \frac{1}{x^2 \sqrt{x^2(1 + \frac{1}{x^2})}} dx = \int \frac{1}{x^3 \sqrt{1 + \frac{1}{x^2}}} dx$.
Let $1 + \frac{1}{x^2} = t^2$.
Then,differentiating both sides with respect to $x$,we get $-\frac{2}{x^3} dx = 2t dt$,which implies $\frac{dx}{x^3} = -t dt$.
Substituting these into the integral: $I = \int -t dt / t = -\int dt = -t + c$.
Since $t = \sqrt{1 + \frac{1}{x^2}} = \sqrt{\frac{x^2+1}{x^2}} = \frac{\sqrt{x^2+1}}{|x|}$,for $x > 0$,$I = -\frac{\sqrt{x^2+1}}{x} + c$.