Integration by substitution Questions in English

(B) Let $I = \int \frac{\sqrt[4]{x}}{\sqrt{x}+\sqrt[4]{x}} dx$.
Substitute $u = \sqrt[4]{x}$,then $x = u^4$ and $dx = 4u^3 du$.
Substituting these into the integral:
$I = \int \frac{u}{u^2+u} (4u^3 du) = 4 \int \frac{u^4}{u(u+1)} du = 4 \int \frac{u^3}{u+1} du$.
Using polynomial division,$\frac{u^3}{u+1} = u^2 - u + 1 - \frac{1}{u+1}$.
$I = 4 \int (u^2 - u + 1 - \frac{1}{u+1}) du = 4 (\frac{u^3}{3} - \frac{u^2}{2} + u - \log|u+1|) + K$.
$I = \frac{4}{3}u^3 - 2u^2 + 4u - 4 \log(u+1) + K$.
Factoring out $\frac{2}{3}$:
$I = \frac{2}{3} [2u^3 - 3u^2 + 6u - 6 \log(u+1)] + K$.
Substituting $u = \sqrt[4]{x}$:
$I = \frac{2}{3} [2 \sqrt[4]{x^3} - 3 \sqrt[4]{x^2} + 6 \sqrt[4]{x} - 6 \log(1+\sqrt[4]{x})] + K$.
Comparing with the given form,we get $A=2, B=-3, C=6, D=-6$.
Thus,$\frac{2}{3}(A+B+C+D) = \frac{2}{3}(2 - 3 + 6 - 6) = \frac{2}{3}(-1) = -\frac{2}{3}$.

(C) Let $I = \int (\log x)^m x^n \, dx$.
Substitute $\log x = t$,which implies $x = e^t$.
Then,differentiating both sides with respect to $t$,we get $dx = e^t \, dt$.
Substituting these into the integral:
$I = \int t^m (e^t)^n \cdot e^t \, dt$
$I = \int t^m e^{nt} \cdot e^t \, dt$
$I = \int t^m e^{(n+1)t} \, dt$.
Thus,the correct substitution is $x = e^t$ and the resulting integral is $\int t^m e^{(n+1)t} \, dt$.

(B) Let $I = \int \frac{dx}{4+5 \cos x}$.
Using the substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and $dx = \frac{2 du}{1+u^2}$ where $u = \tan(x/2)$:
$I = \int \frac{\frac{2 du}{1+u^2}}{4+5\left(\frac{1-u^2}{1+u^2}\right)} = \int \frac{2 du}{4(1+u^2) + 5(1-u^2)} = \int \frac{2 du}{4+4u^2+5-5u^2} = \int \frac{2 du}{9-u^2}$.
Using the formula $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C$:
$I = 2 \times \frac{1}{2(3)} \ln \left| \frac{3+u}{3-u} \right| + C = \frac{1}{3} \ln \left| \frac{3+\tan(x/2)}{3-\tan(x/2)} \right| + C$.

(B) Let $I = \int \sqrt{x}(1-x^3)^{-\frac{1}{2}} dx$.
Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implies $\sqrt{x} dx = \frac{2}{3} dt$.
Since $x^3 = (x^{\frac{3}{2}})^2 = t^2$,the integral becomes:
$I = \int (1-t^2)^{-\frac{1}{2}} \left(\frac{2}{3} dt\right) = \frac{2}{3} \int \frac{1}{\sqrt{1-t^2}} dt$.
Integrating,we get $I = \frac{2}{3} \sin^{-1}(t) + c$.
Substituting $t = x^{\frac{3}{2}}$ back,we have $I = \frac{2}{3} \sin^{-1}(x^{\frac{3}{2}}) + c$.
Comparing this with $\frac{2}{3} g(f(x)) + c$,we identify $f(x) = x^{\frac{3}{2}}$ and $g(x) = \sin^{-1} x$.

(A) Given $\int x^{n-2} f(x) dx = \int x^{n-2} \cdot \frac{x}{(1 + nx^n)^{1/n}} dx = \int \frac{x^{n-1}}{(1 + nx^n)^{1/n}} dx$.
Let $t = 1 + nx^n$.
Then $dt = n \cdot n x^{n-1} dx = n^2 x^{n-1} dx$,which implies $x^{n-1} dx = \frac{1}{n^2} dt$.
Substituting these into the integral:
$\int \frac{1}{n^2} \cdot t^{-1/n} dt = \frac{1}{n^2} \left[ \frac{t^{-1/n + 1}}{-1/n + 1} \right] + C$.
$= \frac{1}{n^2} \left[ \frac{t^{(n-1)/n}}{(n-1)/n} \right] + C = \frac{1}{n^2} \cdot \frac{n}{n-1} \cdot t^{(n-1)/n} + C$.
$= \frac{1}{n(n-1)} (1 + nx^n)^{1 - 1/n} + C$.

(D) Let $I = \int \frac{1+x \cos x}{x[1-x^2(e^{\sin x})^2]} dx$.
Multiply the numerator and denominator by $e^{\sin x}$:
$I = \int \frac{(1+x \cos x) e^{\sin x}}{x e^{\sin x}[1-(x e^{\sin x})^2]} dx$.
Let $t = x e^{\sin x}$.
Then $dt = [e^{\sin x} + x e^{\sin x} \cos x] dx = e^{\sin x}(1+x \cos x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1-t^2)} = \int \frac{dt}{t(1-t)(1+t)}$.
Using partial fractions:
$\frac{1}{t(1-t)(1+t)} = \frac{1}{t} + \frac{1}{2(1-t)} - \frac{1}{2(1+t)}$.
Integrating:
$I = \ln|t| - \frac{1}{2} \ln|1-t| - \frac{1}{2} \ln|1+t| + C = \ln|t| - \frac{1}{2} \ln|1-t^2| + C$.
$I = \frac{1}{2} \ln|t^2| - \frac{1}{2} \ln|1-t^2| + C = \frac{1}{2} \ln \left| \frac{t^2}{1-t^2} \right| + C$.
Substituting back $t = x e^{\sin x}$:
$I = \frac{1}{2} \ln \left| \frac{(x e^{\sin x})^2}{1-(x e^{\sin x})^2} \right| + C = -\frac{1}{2} \ln \left| \frac{(x e^{\sin x})^2}{(x e^{\sin x})^2-1} \right| + C$.

(D) Let $I = \int \frac{d x}{(x-1) \sqrt{x+2}}$.
Substitute $t = \sqrt{x+2}$,which implies $t^2 = x+2$,so $x = t^2 - 2$.
Then $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int \frac{2t \, dt}{(t^2 - 2 - 1)t} = \int \frac{2 \, dt}{t^2 - 3}$.
Using the standard integral formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$:
$I = 2 \times \frac{1}{2\sqrt{3}} \log \left| \frac{t-\sqrt{3}}{t+\sqrt{3}} \right| + C$.
$I = \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}} \right| + C$.

(A) Given $x = \frac{t^3}{t^2 - 1}$ and $y = \frac{t}{t^2 - 1}$.
First,calculate $x - 3y$:
$x - 3y = \frac{t^3}{t^2 - 1} - 3 \left( \frac{t}{t^2 - 1} \right) = \frac{t^3 - 3t}{t^2 - 1}$.
Next,find $dx$ by differentiating $x$ with respect to $t$:
$dx = \frac{d}{dt} \left( \frac{t^3}{t^2 - 1} \right) dt = \frac{(t^2 - 1)(3t^2) - t^3(2t)}{(t^2 - 1)^2} dt = \frac{3t^4 - 3t^2 - 2t^4}{(t^2 - 1)^2} dt = \frac{t^4 - 3t^2}{(t^2 - 1)^2} dt$.
Now,substitute these into the integral:
$\int \frac{dx}{x - 3y} = \int \frac{\frac{t^4 - 3t^2}{(t^2 - 1)^2} dt}{\frac{t^3 - 3t}{t^2 - 1}} = \int \frac{t^2(t^2 - 3)}{(t^2 - 1)^2} \cdot \frac{t^2 - 1}{t(t^2 - 3)} dt = \int \frac{t}{t^2 - 1} dt$.
Let $u = t^2 - 1$,then $du = 2t dt$,so $t dt = \frac{1}{2} du$.
$\int \frac{1}{2u} du = \frac{1}{2} \log|u| + C = \frac{1}{2} \log(t^2 - 1) + C$.

(A) Let $I = \int \frac{d x}{(x-3)^{\frac{4}{5}}(x+1)^{\frac{6}{5}}}$.
We can rewrite the integral as:
$I = \int \frac{d x}{(x+1)^2 \left(\frac{x-3}{x+1}\right)^{\frac{4}{5}}}$.
Let $t = \frac{x-3}{x+1}$.
Then,$dt = \frac{(x+1)(1) - (x-3)(1)}{(x+1)^2} dx = \frac{4}{(x+1)^2} dx$.
Thus,$\frac{dx}{(x+1)^2} = \frac{dt}{4}$.
Substituting these into the integral:
$I = \int \frac{1}{t^{\frac{4}{5}}} \cdot \frac{dt}{4} = \frac{1}{4} \int t^{-\frac{4}{5}} dt$.
$I = \frac{1}{4} \cdot \frac{t^{\frac{1}{5}}}{\frac{1}{5}} + C = \frac{5}{4} t^{\frac{1}{5}} + C$.
Substituting back $t = \frac{x-3}{x+1}$:
$I = \frac{5}{4} \left(\frac{x-3}{x+1}\right)^{\frac{1}{5}} + C$.

(A) Let $I = \int \frac{\sin ^3 x}{(\cos ^4 x + 3 \cos ^2 x + 1) \tan ^{-1}(\sec x + \cos x)} dx$.
Let $t = \tan ^{-1}(\sec x + \cos x)$.
Then $dt = \frac{1}{1 + (\sec x + \cos x)^2} (\sec x \tan x - \sin x) dx$.
$dt = \frac{1}{1 + (\frac{1}{\cos x} + \cos x)^2} (\frac{\sin x}{\cos ^2 x} - \sin x) dx$.
$dt = \frac{\cos ^2 x}{\cos ^2 x + (1 + \cos ^2 x)^2} \cdot \frac{\sin x (1 - \cos ^2 x)}{\cos ^2 x} dx$.
$dt = \frac{\sin x \cdot \sin ^2 x}{\cos ^2 x + 1 + 2 \cos ^2 x + \cos ^4 x} dx = \frac{\sin ^3 x}{\cos ^4 x + 3 \cos ^2 x + 1} dx$.
Thus,$I = \int \frac{dt}{t} = \ln |t| + C = \ln |\tan ^{-1}(\sec x + \cos x)| + C$.
Given $f(x) = \ln |\tan ^{-1}(\sec x + \cos x)|$,we have $e^{f(x)} = \tan ^{-1}(\sec x + \cos x)$.

(B) Given integral $I = \int \frac{1}{(\sin x + \cos x + \sqrt{2} \sqrt{\sin 2x})^2} dx$.
We can rewrite the denominator as $(\sqrt{\sin x} + \sqrt{\cos x})^4$.
Thus,$I = \int \frac{dx}{(\sqrt{\sin x} + \sqrt{\cos x})^4} = \int \frac{dx}{\cos^2 x (\sqrt{\tan x} + 1)^4} = \int \frac{\sec^2 x}{(\sqrt{\tan x} + 1)^4} dx$.
Let $\tan x = t^2$,then $\sec^2 x dx = 2t dt$.
Substituting these into the integral: $I = \int \frac{2t dt}{(t+1)^4} = 2 \int \frac{t+1-1}{(t+1)^4} dt = 2 \int \left( \frac{1}{(t+1)^3} - \frac{1}{(t+1)^4} \right) dt$.
Integrating term by term: $I = 2 \left[ \frac{(t+1)^{-2}}{-2} - \frac{(t+1)^{-3}}{-3} \right] + C = 2 \left[ \frac{1}{3(t+1)^3} - \frac{1}{2(t+1)^2} \right] + C$.
Simplifying: $I = \frac{2}{3(t+1)^3} - \frac{1}{(t+1)^2} + C = \frac{2 - 3(t+1)}{3(t+1)^3} + C = \frac{2 - 3t - 3}{3(t+1)^3} + C = \frac{-(1+3t)}{3(1+t)^3} + C$.
Substituting $t = \sqrt{\tan x}$,we get $I = \frac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$.

(C) Let $I = \int \frac{\cos ^4 x}{\left(\sin ^2 x+\sin ^{-3} x \cos ^5 x\right)^3} d x$.
Factor out $\sin ^2 x$ from the denominator:
$I = \int \frac{\cos ^4 x}{\left(\sin ^2 x(1+\sin ^{-5} x \cos ^5 x)\right)^3} d x = \int \frac{\cos ^4 x}{\sin ^6 x(1+\cot ^5 x)^3} d x$.
This simplifies to:
$I = \int \frac{\cot ^4 x \operatorname{cosec}^2 x}{(1+\cot ^5 x)^3} d x$.
Let $1+\cot ^5 x = y$.
Then $5 \cot ^4 x(-\operatorname{cosec}^2 x) d x = d y$,which implies $\cot ^4 x \operatorname{cosec}^2 x d x = -\frac{1}{5} d y$.
Substituting these into the integral:
$I = -\frac{1}{5} \int y^{-3} d y = -\frac{1}{5} \left(\frac{y^{-2}}{-2}\right) + C = \frac{1}{10} y^{-2} + C$.
Substituting back $y = 1+\cot ^5 x$:
$I = \frac{1}{10}(1+\cot ^5 x)^{-2} + C$.

(B) Given the integral $I = \int \frac{\sin \left(x-\frac{\pi}{4}\right)}{2+\sin 2 x} d x$.
Using the identity $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,the denominator becomes $2 + \sin 2x = 3 - (\sin x - \cos x)^2$. This is not the most direct path.
Alternatively,note that $2 + \sin 2x = 1 + (1 + \sin 2x) = 1 + (\sin x + \cos x)^2$.
Thus,$I = \int \frac{\sin x \cos(\pi/4) - \cos x \sin(\pi/4)}{1 + (\sin x + \cos x)^2} dx = \frac{1}{\sqrt{2}} \int \frac{\sin x - \cos x}{1 + (\sin x + \cos x)^2} dx$.
Let $t = \sin x + \cos x$,then $dt = (\cos x - \sin x) dx$,which implies $-dt = (\sin x - \cos x) dx$.
Substituting these into the integral:
$I = \frac{1}{\sqrt{2}} \int \frac{-dt}{1 + t^2} = -\frac{1}{\sqrt{2}} \tan^{-1}(t) + C$.
Substituting $t = \sin x + \cos x$ back,we get $I = -\frac{1}{\sqrt{2}} \tan^{-1}(\sin x + \cos x) + C$.
Comparing this with the given form $-\frac{1}{\sqrt{2}} \tan^{-1}(f(x)) + C$,we find $f(x) = \sin x + \cos x$.
Since $\sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) = \sqrt{2} \cos(x - \frac{\pi}{4})$,the correct option is $B$.

(A) We have $I = \int \frac{dx}{\sqrt{x^2(1-x^3)}} = \int \frac{dx}{x\sqrt{1-x^3}}$.
Multiply numerator and denominator by $x^2$:
$I = \int \frac{x^2 dx}{x^3\sqrt{1-x^3}}$.
Let $t = \sqrt{1-x^3}$,then $t^2 = 1-x^3$,so $2t dt = -3x^2 dx$,which means $x^2 dx = -\frac{2}{3}t dt$.
Also $x^3 = 1-t^2$.
Substituting these into the integral:
$I = \int \frac{-\frac{2}{3}t dt}{(1-t^2)t} = -\frac{2}{3} \int \frac{dt}{1-t^2} = -\frac{2}{3} \cdot \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + C = \frac{1}{3} \log \left| \frac{1-t}{1+t} \right| + C$.
Substituting $t = \sqrt{1-x^3}$ back:
$I = \frac{1}{3} \log \left| \frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}} \right| + C$.
Thus,$f(x) = \frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}$.
For $x = \frac{1}{2}$,$x^3 = \frac{1}{8}$,so $1-x^3 = \frac{7}{8}$.
$f\left(\frac{1}{2}\right) = \frac{1-\sqrt{7/8}}{1+\sqrt{7/8}} = \frac{\sqrt{8}-\sqrt{7}}{\sqrt{8}+\sqrt{7}}$.

(B) Let $I = \int \frac{dx}{x \sqrt{x^4 - 1}}$.
Multiply the numerator and denominator by $x$:
$I = \int \frac{x \ dx}{x^2 \sqrt{(x^2)^2 - 1}}$.
Let $x^2 = t$,then $2x \ dx = dt$,or $x \ dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{dt/2}{t \sqrt{t^2 - 1}} = \frac{1}{2} \int \frac{dt}{t \sqrt{t^2 - 1}}$.
We know that $\int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + C$.
Therefore,$I = \frac{1}{2} \sec^{-1}(t) + C = \frac{1}{2} \sec^{-1}(x^2) + C$.
Comparing this with the given expression $\frac{1}{k} \sec^{-1}(x^k)$,we get $k = 2$.

(C) We are given $f(x) = \frac{1}{\cos^2 x \sqrt{1 + \tan x}} = \sec^2 x (1 + \tan x)^{-1/2}$.
Let $t = 1 + \tan x$. Then $dt = \sec^2 x \ dx$.
Substituting these into the integral:
$F(x) = \int f(x) \ dx = \int (1 + \tan x)^{-1/2} \sec^2 x \ dx = \int t^{-1/2} \ dt$.
Integrating with respect to $t$:
$F(x) = \frac{t^{1/2}}{1/2} + C = 2 \sqrt{t} + C = 2 \sqrt{1 + \tan x} + C$.
Given $F(0) = 4$,we substitute $x = 0$:
$F(0) = 2 \sqrt{1 + \tan 0} + C = 2 \sqrt{1 + 0} + C = 2 + C$.
Since $F(0) = 4$,we have $2 + C = 4$,which implies $C = 2$.
Thus,$F(x) = 2 \sqrt{1 + \tan x} + 2 = 2 (\sqrt{1 + \tan x} + 1)$.

(B) Given $I = \int \frac{\sec x}{\sqrt{\sin (2 x + \theta) + \sin \theta}} d x$
Using the formula $\sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2})$:
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \sin (\frac{2 x + 2 \theta}{2}) \cos (\frac{2 x + \theta - \theta}{2})}} d x$
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \sin (x + \theta) \cos x}} d x$
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 (\sin x \cos \theta + \cos x \sin \theta) \cos x}} d x$
Divide numerator and denominator by $\cos x$ inside the square root:
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \cos^2 x (\tan x \cos \theta + \sin \theta)}} d x$
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2} \cos x \sqrt{\tan x \cos \theta + \sin \theta}} d x$
$\Rightarrow I = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta + \sin \theta}} d x$
Let $t = \tan x \cos \theta + \sin \theta$,then $dt = \sec^2 x \cos \theta d x$,so $\sec^2 x d x = \frac{dt}{\cos \theta}$.
$\Rightarrow I = \frac{1}{\sqrt{2} \cos \theta} \int t^{-1/2} dt$
$\Rightarrow I = \frac{1}{\sqrt{2} \cos \theta} \cdot 2 t^{1/2} + C$
$\Rightarrow I = \sqrt{\frac{2}{\cos^2 \theta}} \sqrt{\tan x \cos \theta + \sin \theta} + C$
$\Rightarrow I = \sqrt{2 \sec^2 \theta (\tan x \cos \theta + \sin \theta)} + C$
$\Rightarrow I = \sqrt{2 \sec \theta (\tan x + \tan \theta)} + C$

(B) Given the integral: $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.
Let $I = \int \cos ^k(x) \sin (x) d x$.
Substitute $t = \cos x$,then $dt = -\sin x d x$,which implies $\sin x d x = -dt$.
Substituting these into the integral,we get:
$I = \int t^k (-dt) = -\int t^k dt$.
Integrating $t^k$ with respect to $t$ gives:
$I = -\frac{t^{k+1}}{k+1} + C$.
Substituting back $t = \cos x$:
$I = -\frac{\cos^{k+1} x}{k+1} + C$.
Comparing this with the given expression $\frac{-1}{4} \cos^4(x) + C$,we have:
$\frac{1}{k+1} = \frac{1}{4}$ and $k+1 = 4$.
Therefore,$k = 3$.

(C) Let $I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 1 - (\sin x - \cos x)^2$.
Alternatively,we can write $\sin 2x = (\sin x + \cos x)^2 - 1$.
So,$I = \int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^2 - 1}} \, dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) \, dx$,which implies $(\sin x - \cos x) \, dx = -dt$.
Substituting these into the integral:
$I = \int \frac{-dt}{\sqrt{t^2 - 1}} = -\int \frac{dt}{\sqrt{t^2 - 1}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + C$,we get:
$I = -\log |t + \sqrt{t^2 - 1}| + C$.
Substituting $t = \sin x + \cos x$ back:
$I = -\log |\sin x + \cos x + \sqrt{(\sin x + \cos x)^2 - 1}| + C$.
Since $(\sin x + \cos x)^2 - 1 = 1 + \sin 2x - 1 = \sin 2x$,we have:
$I = -\log |\sin x + \cos x + \sqrt{\sin 2x}| + C$.

(B) Let $I = \int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}\right)^{\frac{1}{7}} d x = \int x^{\frac{1}{3}}\left(1+x^{\frac{4}{3}}\right)^{\frac{1}{7}} d x$.
Let $1+x^{\frac{4}{3}} = t$.
Then,$\frac{4}{3} x^{\frac{1}{3}} d x = d t$,which implies $x^{\frac{1}{3}} d x = \frac{3}{4} d t$.
Substituting these into the integral,we get:
$I = \int t^{\frac{1}{7}} \cdot \frac{3}{4} d t = \frac{3}{4} \int t^{\frac{1}{7}} d t$.
Integrating,we have:
$I = \frac{3}{4} \cdot \frac{t^{\frac{1}{7}+1}}{\frac{1}{7}+1} + c = \frac{3}{4} \cdot \frac{t^{\frac{8}{7}}}{\frac{8}{7}} + c = \frac{3}{4} \cdot \frac{7}{8} t^{\frac{8}{7}} + c = \frac{21}{32} \left(1+x^{\frac{4}{3}}\right)^{\frac{8}{7}} + c$.
Comparing this with $A\left(1+\sqrt[3]{x^4}\right)^B+c$,we get $A = \frac{21}{32}$ and $B = \frac{8}{7}$.
Therefore,$A B = \frac{21}{32} \times \frac{8}{7} = \frac{3}{4}$.

(D) Let $I = \int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$.
Using the substitution $t = x+\frac{\pi}{4}$,we have $x = t-\frac{\pi}{4}$ and $dx = dt$.
Substituting these into the integral:
$I = \int \frac{\sqrt{2} \sin \left(t-\frac{\pi}{4}\right)}{\sin t} dt$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \int \sqrt{2} \frac{\sin t \cos \frac{\pi}{4} - \cos t \sin \frac{\pi}{4}}{\sin t} dt$.
Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$I = \int \sqrt{2} \frac{\left(\frac{1}{\sqrt{2}} \sin t - \frac{1}{\sqrt{2}} \cos t\right)}{\sin t} dt = \int \frac{\sin t - \cos t}{\sin t} dt$.
$I = \int (1 - \cot t) dt = t - \ln |\sin t| + c$.
Substituting back $t = x+\frac{\pi}{4}$:
$I = x + \frac{\pi}{4} - \ln |\sin (x+\frac{\pi}{4})| + c$.
Since $\frac{\pi}{4}$ is a constant,it can be absorbed into the constant $c$:
$I = x - \log |\sin (x+\frac{\pi}{4})| + c$.

(C) We have,$\int \frac{2x+3}{x(x+1)(x+2)(x+3)+1} dx = \frac{-1}{ax^2+bx+c} + \alpha$.
Rearranging the denominator: $x(x+3)(x+1)(x+2)+1 = (x^2+3x)(x^2+3x+2)+1$.
Let $t = x^2+3x$,then $dt = (2x+3)dx$.
Substituting into the integral: $\int \frac{dt}{t(t+2)+1} = \int \frac{dt}{t^2+2t+1} = \int \frac{dt}{(t+1)^2}$.
Integrating gives: $-\frac{1}{t+1} + \alpha = \frac{-1}{x^2+3x+1} + \alpha$.
Comparing this with $\frac{-1}{ax^2+bx+c} + \alpha$,we get $a=1, b=3, c=1$.
Therefore,$a+b+c = 1+3+1 = 5$.

(B) We have $I = \int \frac{2 \tan x}{1+2 \tan^2 x} dx$.
Converting to $\sin x$ and $\cos x$:
$I = \int \frac{2 \frac{\sin x}{\cos x}}{1+2 \frac{\sin^2 x}{\cos^2 x}} dx = \int \frac{2 \sin x \cos x}{\cos^2 x + 2 \sin^2 x} dx$.
Using $\cos^2 x = 1 - \sin^2 x$,the denominator becomes $(1 - \sin^2 x) + 2 \sin^2 x = 1 + \sin^2 x$.
So,$I = \int \frac{2 \sin x \cos x}{1 + \sin^2 x} dx$.
Let $t = 1 + \sin^2 x$. Then $dt = 2 \sin x \cos x dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t} = \log |t| + c_1 = \log |1 + \sin^2 x| + c_1$.
Since $1 = \cos^2 x + \sin^2 x$,we have $1 + \sin^2 x = \cos^2 x + 2 \sin^2 x$.
$I = \log |\cos^2 x + 2 \sin^2 x| + c_1 = \log |2(\frac{\cos^2 x}{2} + \sin^2 x)| + c_1$.
Using $\log(ab) = \log a + \log b$:
$I = \log 2 + \log |\frac{\cos^2 x}{2} + \sin^2 x| + c_1 = \log |\frac{\cos^2 x}{2} + \sin^2 x| + c$,where $c = c_1 + \log 2$.

(D) Let $I = \int \frac{\cos^3 x}{\sin^2 x + \sin x} \, dx$.
Using the identity $\cos^2 x = 1 - \sin^2 x$,we can rewrite the integral as:
$I = \int \frac{(1 - \sin^2 x) \cos x}{\sin^2 x + \sin x} \, dx$.
Substitute $\sin x = t$,which implies $\cos x \, dx = dt$.
Substituting these into the integral:
$I = \int \frac{1 - t^2}{t^2 + t} \, dt = \int \frac{(1 - t)(1 + t)}{t(t + 1)} \, dt$.
Canceling the common term $(1 + t)$:
$I = \int \frac{1 - t}{t} \, dt = \int \left( \frac{1}{t} - 1 \right) \, dt$.
Integrating with respect to $t$:
$I = \log |t| - t + C$.
Substituting back $t = \sin x$:
$I = \log |\sin x| - \sin x + C$.

(D) Let $I = \int \sin (\sqrt{k}) \, dk$ for $k \in (0, \infty)$.
Substitute $k = t^2$,which implies $dk = 2t \, dt$.
Substituting these into the integral,we get:
$I = \int \sin(t) \cdot (2t \, dt) = 2 \int t \sin(t) \, dt$.
Using integration by parts,where $\int u \, dv = uv - \int v \, du$,let $u = t$ and $dv = \sin(t) \, dt$. Then $du = dt$ and $v = -\cos(t)$.
$I = 2 [t(-\cos(t)) - \int (-\cos(t)) \, dt]$
$I = 2 [-t \cos(t) + \int \cos(t) \, dt]$
$I = 2 [-t \cos(t) + \sin(t)] + c$
Substituting $t = \sqrt{k}$ back into the expression:
$I = 2[\sin(\sqrt{k}) - \sqrt{k} \cos(\sqrt{k})] + c$.

(B) Given the equation of the curve is $y = \int x^3 e^{x^4} dx$.
Let $t = x^4$,then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$.
Substituting this into the integral,we get $y = \frac{1}{4} \int e^t dt = \frac{1}{4} e^t + C = \frac{1}{4} e^{x^4} + C$.
Since the curve passes through $(0,1)$,we substitute $x=0$ and $y=1$: $1 = \frac{1}{4} e^0 + C \Rightarrow 1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$.
Thus,the equation is $y = \frac{1}{4} e^{x^4} + \frac{3}{4}$.
Rearranging for $x$: $y - \frac{3}{4} = \frac{1}{4} e^{x^4} \Rightarrow 4y - 3 = e^{x^4}$.
Taking the natural logarithm on both sides: $\log_e |4y - 3| = x^4$.
Therefore,$x = (\log_e |4y - 3|)^{1/4}$.
Thus,$f(y) = (\log_e |4y - 3|)^{1/4}$.

(B) Let $I = \int (1+e^{-x})^{-1} dx = \int \frac{1}{1 + \frac{1}{e^x}} dx$
$I = \int \frac{e^x}{e^x+1} dx$
Let $u = e^x + 1$,then $du = e^x dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{u} du = \log |u| + C$
Substituting back $u = e^x + 1$:
$I = \log (e^x + 1) + C$.

(B) Let $I = \int \frac{\cos x-\sin x}{5+\sin 2 x} d x$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$ is not helpful here,but $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x$.
Thus,$\sin 2x = (\sin x + \cos x)^2 - 1$.
Substituting this into the integral:
$I = \int \frac{\cos x-\sin x}{5 + (\sin x + \cos x)^2 - 1} d x = \int \frac{\cos x-\sin x}{4 + (\sin x + \cos x)^2} d x$.
Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$.
Substituting $t$ and $dt$ into the integral:
$I = \int \frac{dt}{4 + t^2} = \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) + C$.
Substituting back $t = \sin x + \cos x$:
$I = \frac{1}{2} \tan^{-1}\left(\frac{\sin x + \cos x}{2}\right) + C$.

(B) Given the integral $I = \int e^{3 \log x}(x^4+1)^{-1} dx$.
Using the property $e^{\log a} = a$,we have $e^{3 \log x} = e^{\log x^3} = x^3$.
So,$I = \int \frac{x^3}{x^4+1} dx$.
Let $t = x^4+1$.
Then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{4t} dt = \frac{1}{4} \log|t| + C$.
Substituting back $t = x^4+1$,we obtain $I = \frac{1}{4} \log(x^4+1) + C$.

(D) Let $I = \int \frac{\sin(2x)}{\sin^2(x) + 2\cos^2(x)} dx$.
Using the identity $\sin^2(x) = 1 - \cos^2(x)$,the denominator becomes $1 - \cos^2(x) + 2\cos^2(x) = 1 + \cos^2(x)$.
So,$I = \int \frac{\sin(2x)}{1 + \cos^2(x)} dx$.
Let $t = 1 + \cos^2(x)$.
Then $dt = 2\cos(x)(-\sin(x)) dx = -\sin(2x) dx$.
This implies $\sin(2x) dx = -dt$.
Substituting these into the integral:
$I = \int \frac{-dt}{t} = -\log |t| + c$.
Substituting back $t = 1 + \cos^2(x)$,we get $I = -\log |1 + \cos^2(x)| + c$.
Thus,option $D$ is correct.

(A) Let $I = \int \frac{x^{n-1}}{x^{2n} + 4} dx$.
Substitute $x^n = t$.
Then,differentiating both sides with respect to $x$,we get $n x^{n-1} dx = dt$,which implies $x^{n-1} dx = \frac{1}{n} dt$.
Substituting these into the integral,we get:
$I = \frac{1}{n} \int \frac{dt}{t^2 + 2^2}$.
Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + c$,we have:
$I = \frac{1}{n} \cdot \frac{1}{2} \tan^{-1} \left( \frac{t}{2} \right) + c$.
Substituting $t = x^n$ back,we get:
$I = \frac{1}{2n} \tan^{-1} \left( \frac{x^n}{2} \right) + c$.

(D) Let $I = \int \frac{1 + \tan^{2} x}{1 - \tan^{2} x} dx$.
Since $1 + \tan^{2} x = \sec^{2} x$,we have $I = \int \frac{\sec^{2} x}{1 - \tan^{2} x} dx$.
Substitute $\tan x = t$,then $\sec^{2} x \ dx = dt$.
Substituting these into the integral,we get $I = \int \frac{dt}{1 - t^{2}}$.
Using the standard integral formula $\int \frac{dx}{a^{2} - x^{2}} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + c$,we get $I = \frac{1}{2} \log \left| \frac{1 + t}{1 - t} \right| + c$.
Replacing $t$ with $\tan x$,we obtain $I = \frac{1}{2} \log \left| \frac{1 + \tan x}{1 - \tan x} \right| + c$.

(A) Let $I = \int \frac{d x}{\sqrt{2 e^x-1}}$.
Substitute $u = \sqrt{2 e^x-1}$. Then $u^2 = 2 e^x - 1$,which implies $2 e^x = u^2 + 1$,or $e^x = \frac{u^2+1}{2}$.
Taking the natural logarithm on both sides,$x = \ln\left(\frac{u^2+1}{2}\right)$.
Differentiating with respect to $u$,$dx = \frac{1}{\frac{u^2+1}{2}} \cdot \frac{2u}{2} du = \frac{2u}{u^2+1} du$.
Substituting these into the integral:
$I = \int \frac{1}{u} \cdot \frac{2u}{u^2+1} du = \int \frac{2}{u^2+1} du$.
Integrating,we get $I = 2 \operatorname{Tan}^{-1}(u) + c$.
Substituting back $u = \sqrt{2 e^x-1}$,we get $I = 2 \operatorname{Tan}^{-1}\left(\sqrt{2 e^x-1}\right) + c$.

(A) Given integral is $I = \int \frac{dx}{e^x + 4e^{-x}}$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{e^x dx}{e^{2x} + 4}$.
Let $u = e^x$,then $du = e^x dx$.
The integral becomes $I = \int \frac{du}{u^2 + 2^2}$.
Using the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{2} \tan^{-1}(\frac{u}{2}) + c$.
Substituting $u = e^x$ back,we get $I = \frac{1}{2} \tan^{-1}(\frac{e^x}{2}) + c$.
Thus,$f(x) = \frac{1}{2} \tan^{-1}(\frac{e^x}{2})$.

(A) To solve the integral $\int \frac{2 t+1}{t^2+t+1} d t$,we use the method of substitution.
Let $u = t^2+t+1$.
Then,the derivative of $u$ with respect to $t$ is $\frac{du}{dt} = 2t+1$.
This implies $du = (2t+1) dt$.
Substituting these into the original integral,we get $\int \frac{1}{u} du$.
The integral of $\frac{1}{u}$ is $\log |u| + c$.
Substituting back $u = t^2+t+1$,we obtain $\log |t^2+t+1| + c$.

(C) Let $I = \int \sin ^5 x \cos ^5 x \, dx$.
We can rewrite the integral as:
$I = \int \cos ^5 x \sin ^4 x \sin x \, dx = \int \cos ^5 x (1 - \cos ^2 x)^2 \sin x \, dx$.
Let $\cos x = t$,then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Substituting these into the integral:
$I = \int t^5 (1 - t^2)^2 (-dt) = -\int t^5 (t^4 - 2t^2 + 1) \, dt$.
$I = -\int (t^9 - 2t^7 + t^5) \, dt = -(\frac{t^{10}}{10} - 2 \frac{t^8}{8} + \frac{t^6}{6}) + C$.
$I = -\frac{t^6}{60} (6t^4 - 15t^2 + 10) + C$.
Substituting $t = \cos x$:
$I = -\frac{\cos ^6 x}{60} (6 \cos ^4 x - 15 \cos ^2 x + 10) + C$.
Using $\cos ^2 x = 1 - \sin ^2 x$:
$I = -\frac{\cos ^6 x}{60} [6(1 - \sin ^2 x)^2 - 15(1 - \sin ^2 x) + 10] + C$.
$I = -\frac{\cos ^6 x}{60} [6(1 - 2 \sin ^2 x + \sin ^4 x) - 15 + 15 \sin ^2 x + 10] + C$.
$I = -\frac{\cos ^6 x}{60} [6 - 12 \sin ^2 x + 6 \sin ^4 x - 15 + 15 \sin ^2 x + 10] + C$.
$I = -\frac{\cos ^6 x}{60} [6 \sin ^4 x + 3 \sin ^2 x + 1] + C$.
Thus,option $(c)$ is correct.

(D) Given $I = \int \frac{x-x^2}{\sqrt{1-x}} d x$.
Since $x-x^2 = x(1-x)$,we have $I = \int \frac{x(1-x)}{\sqrt{1-x}} d x = \int x \sqrt{1-x} d x$.
Let $1-x = t^2$,then $dx = -2t dt$.
Substituting these into the integral:
$I = \int (1-t^2) t (-2t) dt = 2 \int (t^4 - t^2) dt$.
Integrating with respect to $t$:
$I = 2 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right] + c = \frac{2}{15} t^3 (3t^2 - 5) + c$.
Substituting $t = \sqrt{1-x}$ back:
$I = \frac{2}{15} (1-x)^{3/2} [3(1-x) - 5] + c = \frac{2}{15} (1-x)^{3/2} (-3x - 2) + c = -\frac{2}{15} (1-x)^{3/2} (3x+2) + c$.

(D) Given the integral $I = \int \frac{dx}{(2 \sin x + \sec x)^4}$.
Multiply numerator and denominator by $\sec^4 x$:
$I = \int \frac{\sec^4 x}{(2 \sin x \cos x + 1)^4} dx = \int \frac{\sec^4 x}{(\sin 2x + 1)^4} dx$.
Alternatively,rewrite the denominator: $2 \sin x + \sec x = \frac{2 \sin x \cos x + 1}{\cos x} = \frac{\sin 2x + 1}{\cos x}$.
Thus,$I = \int \frac{\cos^4 x}{(\sin 2x + 1)^4} dx = \int \frac{\cos^4 x}{((\sin x + \cos x)^2)^4} dx = \int \frac{\cos^4 x}{(\sin x + \cos x)^8} dx$.
Divide numerator and denominator by $\cos^8 x$:
$I = \int \frac{\sec^4 x}{(1 + \tan x)^8} dx$.
Let $\tan x = t$,then $\sec^2 x dx = dt$. Since $\sec^2 x = 1 + t^2$,we have:
$I = \int \frac{1 + t^2}{(1 + t)^8} dt = \int \frac{(1 + t)^2 - 2t}{(1 + t)^8} dt = \int \frac{(1 + t)^2 - 2(1 + t - 1)}{(1 + t)^8} dt$
$I = \int \frac{dt}{(1 + t)^6} - 2 \int \frac{dt}{(1 + t)^7} + 2 \int \frac{dt}{(1 + t)^8}$
$I = -\frac{1}{5}(1 + t)^{-5} + \frac{2}{6}(1 + t)^{-6} - \frac{2}{7}(1 + t)^{-7} + k$
$I = -\frac{1}{5}(1 + \tan x)^{-5} + \frac{1}{3}(1 + \tan x)^{-6} - \frac{2}{7}(1 + \tan x)^{-7} + k$.
Comparing with the given form,$A = -\frac{1}{5}$,$B = \frac{1}{3}$,$C = -\frac{2}{7}$.
$A + B + C = -\frac{1}{5} + \frac{1}{3} - \frac{2}{7} = \frac{-21 + 35 - 30}{105} = -\frac{16}{105}$.

(C) Let $I = \int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx$.
Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implies $x^{\frac{1}{2}} dx = \frac{2}{3} dt$.
Also,$t^2 = (x^{\frac{3}{2}})^2 = x^3$.
Substituting these into the integral:
$I = \int \frac{\frac{2}{3} dt}{\sqrt{a^3 - t^2}} = \frac{2}{3} \int \frac{dt}{\sqrt{(\sqrt{a^3})^2 - t^2}}$.
Using the standard integral formula $\int \frac{du}{\sqrt{A^2 - u^2}} = \sin^{-1}(\frac{u}{A}) + c$:
$I = \frac{2}{3} \sin^{-1}\left(\frac{t}{\sqrt{a^3}}\right) + c$.
Substituting $t = x^{\frac{3}{2}}$ back:
$I = \frac{2}{3} \sin^{-1}\left(\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}\right) + c = \frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right) + c$.
Thus,$P(x) = \frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right)$.

(C) We have the integral $I = \int \frac{1-x^7}{x(1+x^7)} dx$.
Multiply the numerator and denominator by $x^6$ to simplify the expression:
$I = \int \frac{x^6(1-x^7)}{x^7(1+x^7)} dx$.
Let $u = x^7$,then $du = 7x^6 dx$,which implies $x^6 dx = \frac{du}{7}$.
Substituting these into the integral:
$I = \int \frac{1-u}{u(1+u)} \cdot \frac{du}{7} = \frac{1}{7} \int \frac{1-u}{u(1+u)} du$.
Using partial fractions: $\frac{1-u}{u(1+u)} = \frac{A}{u} + \frac{B}{1+u}$.
$1-u = A(1+u) + Bu$.
For $u=0$,$A=1$.
For $u=-1$,$1-(-1) = B(-1) \Rightarrow B = -2$.
So,$I = \frac{1}{7} \int (\frac{1}{u} - \frac{2}{1+u}) du = \frac{1}{7} (\ln |u| - 2 \ln |1+u|) + C$.
Substituting $u = x^7$ back:
$I = \frac{1}{7} \ln |x^7| - \frac{2}{7} \ln |x^7+1| + C = \frac{7}{7} \ln |x| - \frac{2}{7} \ln |x^7+1| + C = 1 \ln |x| - \frac{2}{7} \ln |x^7+1| + C$.
Comparing this with $a \ln |x| + b \ln |x^7+1| + c$,we get $a = 1$ and $b = -2/7$.

(C) Let $I = \int \frac{d x}{x^{2 / 3}(1+x^{2 / 3})}$.
Substitute $x^{1/3} = t$. Then $x = t^3$,which implies $dx = 3t^2 dt$.
Substituting these into the integral:
$I = \int \frac{3t^2 dt}{(t^2)(1+t^2)} = \int \frac{3 dt}{1+t^2}$.
The integral of $\frac{1}{1+t^2}$ is $\tan^{-1}(t)$.
Therefore,$I = 3 \tan^{-1}(t) + c$.
Substituting back $t = x^{1/3}$,we get $I = 3 \tan^{-1}(x^{1/3}) + c$.

(C) Let $I = \int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} dx$ ... $(i)$
Put $x^e + e^x = t$ ... $(ii)$
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(x^e + e^x) = \frac{dt}{dx}$
$e x^{e-1} + e^x = \frac{dt}{dx}$
$(e x^{e-1} + e^x) dx = dt$
Factor out $e$:
$e(x^{e-1} + \frac{e^x}{e}) dx = dt$
$e(x^{e-1} + e^{x-1}) dx = dt$
$(x^{e-1} + e^{x-1}) dx = \frac{1}{e} dt$ ... $(iii)$
Substituting $(ii)$ and $(iii)$ into $(i)$:
$I = \int \frac{1}{t} \cdot \frac{1}{e} dt$
$I = \frac{1}{e} \int \frac{1}{t} dt$
$I = \frac{1}{e} \log |t| + C$
Substituting back $t = x^e + e^x$:
$I = \frac{1}{e} \log |x^e + e^x| + C$

(B) Let $I = \int \frac{x+1}{x(1+x e^x)} d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)} d x$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) d x = e^x(1+x) d x$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{t(1+t)}$.
Using partial fractions:
$\frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}$.
Integrating both terms:
$I = \int \left( \frac{1}{t} - \frac{1}{1+t} \right) dt = \log |t| - \log |1+t| + C = \log \left| \frac{t}{1+t} \right| + C$.
Substituting $t = x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + C$.

(B) Let $I = \int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)} [\log g(x)-\log f(x)] \, dx$.
We know that $\frac{d}{dx} \left( \log \frac{g(x)}{f(x)} \right) = \frac{d}{dx} (\log g(x) - \log f(x)) = \frac{g^{\prime}(x)}{g(x)} - \frac{f^{\prime}(x)}{f(x)} = \frac{f(x) g^{\prime}(x) - f^{\prime}(x) g(x)}{f(x) g(x)}$.
Thus,the integral becomes $I = \int \log \left[\frac{g(x)}{f(x)}\right] \cdot d\left( \log \frac{g(x)}{f(x)} \right)$.
Let $t = \log \left[\frac{g(x)}{f(x)}\right]$,then $dt = d\left( \log \frac{g(x)}{f(x)} \right)$.
Substituting this into the integral,we get $I = \int t \, dt = \frac{t^2}{2} + C$.
Substituting back the value of $t$,we get $I = \frac{1}{2} \left[ \log \frac{g(x)}{f(x)} \right]^2 + C$.

(B) Given,$\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$.
We can rewrite the integral as:
$\int \frac{d x}{\sqrt{\sin ^4 x \cdot \frac{\cos x}{\sin x}}} = \int \frac{d x}{\sin ^2 x \sqrt{\cot x}}$
This simplifies to:
$\int \operatorname{cosec}^2 x \cdot (\cot x)^{-1/2} d x$
Let $t = \cot x$,then $dt = -\operatorname{cosec}^2 x d x$,which implies $\operatorname{cosec}^2 x d x = -dt$.
Substituting these into the integral:
$\int -t^{-1/2} dt = -\frac{t^{1/2}}{1/2} + c = -2\sqrt{t} + c$
Substituting back $t = \cot x$:
$-2\sqrt{\cot x} + c = -\frac{2}{\sqrt{\tan x}} + c$
Therefore,$g(x) = -\frac{2}{\sqrt{\tan x}}$.

(C) Let $I = \int \frac{dx}{x^2 \sqrt{4+x^2}}$.
Substitute $x = 2 \tan \theta$,then $dx = 2 \sec^2 \theta \ d\theta$.
The integral becomes $I = \int \frac{2 \sec^2 \theta \ d\theta}{(4 \tan^2 \theta) \sqrt{4 + 4 \tan^2 \theta}}$.
Since $\sqrt{4(1 + \tan^2 \theta)} = 2 \sec \theta$,we have $I = \int \frac{2 \sec^2 \theta \ d\theta}{4 \tan^2 \theta \cdot 2 \sec \theta} = \frac{1}{4} \int \frac{\sec \theta}{\tan^2 \theta} \ d\theta$.
$I = \frac{1}{4} \int \frac{1/\cos \theta}{\sin^2 \theta / \cos^2 \theta} \ d\theta = \frac{1}{4} \int \frac{\cos \theta}{\sin^2 \theta} \ d\theta$.
Let $u = \sin \theta$,then $du = \cos \theta \ d\theta$. So $I = \frac{1}{4} \int u^{-2} \ du = \frac{1}{4} (-u^{-1}) + C = -\frac{1}{4 \sin \theta} + C$.
Since $\tan \theta = \frac{x}{2}$,we have $\sin \theta = \frac{x}{\sqrt{x^2+4}}$.
Thus,$I = -\frac{1}{4} \cdot \frac{\sqrt{x^2+4}}{x} + C = -\frac{\sqrt{4+x^2}}{4x} + C$.

(A) Let $I = \int \frac{dx}{\sqrt{x-x^2}}$.
We can rewrite the denominator as $\sqrt{x(1-x)}$.
So,$I = \int \frac{dx}{\sqrt{x} \sqrt{1-x}}$.
Let $\sqrt{x} = \sin \theta$. Then $x = \sin^2 \theta$.
Differentiating both sides with respect to $\theta$,we get $dx = 2 \sin \theta \cos \theta \, d\theta$.
Substituting these into the integral:
$I = \int \frac{2 \sin \theta \cos \theta \, d\theta}{\sin \theta \sqrt{1-\sin^2 \theta}}$
$I = \int \frac{2 \sin \theta \cos \theta \, d\theta}{\sin \theta \cos \theta}$
$I = \int 2 \, d\theta = 2\theta + C$.
Since $\sin \theta = \sqrt{x}$,we have $\theta = \sin^{-1} \sqrt{x}$.
Therefore,$I = 2 \sin^{-1} \sqrt{x} + C$.