Integration by substitution Questions in English

(A) Let $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $.
Using $\sin 2x = 2\sin x \cos x$,we get $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {4\sin x \cos x} }}} = \frac{1}{2}\int {\frac{{dx}}{{{{\cos }^{7/2}}x{{\sin }^{1/2}}x}}} $.
This can be written as $I = \frac{1}{2}\int {\frac{{{{\sec }^4}x}}{{\sqrt {\tan x} }}dx} = \frac{1}{2}\int {\frac{{(1 + {{\tan }^2}x){{\sec }^2}x}}{{\sqrt {\tan x} }}dx} $.
Substitute $\tan x = t$,so ${\sec ^2}x dx = dt$.
$I = \frac{1}{2}\int {\frac{{1 + {t^2}}}{{\sqrt t }}dt} = \frac{1}{2}\int {({t^{ - 1/2}} + {t^{3/2}})dt} $.
$I = \frac{1}{2} [2{t^{1/2}} + \frac{2}{5}{t^{5/2}}] + c = {t^{1/2}} + \frac{1}{5}{t^{5/2}} + c$.
Substituting back $t = \tan x$,we get $I = \sqrt {\tan x} + \frac{1}{5}{\tan ^{5/2}}x + c$.

(A) Let $I = \int \frac{3^x}{\sqrt{9^x - 1}} \, dx$.
We can rewrite the integral as $I = \int \frac{3^x}{\sqrt{(3^x)^2 - 1}} \, dx$.
Let $3^x = t$. 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 $\int \frac{dt}{\sqrt{t^2 - a^2}} = \log |t + \sqrt{t^2 - a^2}| + c$,we obtain $I = \frac{1}{\log 3} \log |t + \sqrt{t^2 - 1}| + c$.
Finally,substituting $t = 3^x$ back,we get $I = \frac{1}{\log 3} \log |3^x + \sqrt{9^x - 1}| + c$.

(A) Let $I = \int \frac{e^x dx}{\sqrt{a + b e^x}}$.
Substitute $a + b e^x = t$. Then,differentiating both sides with respect to $x$,we get $b e^x dx = dt$,which implies $e^x dx = \frac{1}{b} dt$.
Substituting these into the integral,we get:
$I = \int \frac{1}{\sqrt{t}} \cdot \frac{1}{b} dt = \frac{1}{b} \int t^{-1/2} dt$.
Using the power rule for integration,$\int t^n dt = \frac{t^{n+1}}{n+1} + c$,we have:
$I = \frac{1}{b} \cdot \frac{t^{1/2}}{1/2} + c = \frac{2}{b} \sqrt{t} + c$.
Substituting $t = a + b e^x$ back,we get:
$I = \frac{2}{b} \sqrt{a + b e^x} + c$.

(A) Let $t = 1 + \ln |x|$. Then $dt = \frac{1}{x} \, dx$,and $\ln |x| = t - 1$.
Substituting these into the integral:
$\int \frac{t - 1}{\sqrt{t}} \, dt = \int (\sqrt{t} - \frac{1}{\sqrt{t}}) \, dt$
$= \int (t^{1/2} - t^{-1/2}) \, dt$
$= \frac{t^{3/2}}{3/2} - \frac{t^{1/2}}{1/2} + c$
$= \frac{2}{3} t^{3/2} - 2 t^{1/2} + c$
$= \frac{2}{3} \sqrt{t} (t - 3) + c$
Substituting $t = 1 + \ln |x|$ back:
$= \frac{2}{3} \sqrt{1 + \ln |x|} (1 + \ln |x| - 3) + c$
$= \frac{2}{3} \sqrt{1 + \ln |x|} (\ln |x| - 2) + c$.

(C) Let $I = \int x \cdot 2^{\ln(x^2 + 1)} dx$.
Substitute $t = x^2 + 1$,then $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.
The integral becomes $I = \frac{1}{2} \int 2^{\ln t} dt$.
Using the property $a^{\ln b} = b^{\ln a}$,we have $2^{\ln t} = t^{\ln 2}$.
So,$I = \frac{1}{2} \int t^{\ln 2} dt$.
Integrating with respect to $t$,we get $I = \frac{1}{2} \cdot \frac{t^{\ln 2 + 1}}{\ln 2 + 1} + C$.
Substituting back $t = x^2 + 1$,we get $I = \frac{(x^2 + 1)^{\ln 2 + 1}}{2(\ln 2 + 1)} + C$.

(B) Let $I = \int \frac{\ln(6x^2)}{x} \, dx$.
Substitute $t = \ln(6x^2)$.
Then,$dt = \frac{1}{6x^2} \cdot \frac{d}{dx}(6x^2) \, dx = \frac{1}{6x^2} \cdot 12x \, dx = \frac{2}{x} \, dx$.
This implies $\frac{dx}{x} = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int t \cdot \frac{dt}{2} = \frac{1}{2} \int t \, dt$.
Integrating with respect to $t$:
$I = \frac{1}{2} \cdot \frac{t^2}{2} + C = \frac{t^2}{4} + C$.
Substituting $t = \ln(6x^2)$ back:
$I = \frac{1}{4} [\ln(6x^2)]^2 + C$.

(C) Let $t = \sin x$,then $dt = \cos x \, dx$. The integral becomes:
$I = \int \frac{\cos^2 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x \, dx = \int \frac{(1 - t^2)(1 + 1 - t^2)}{t^2(1 + t^2)} \, dt = \int \frac{(1 - t^2)(2 - t^2)}{t^2(1 + t^2)} \, dt$.
Let $y = t^2$,then $dy = 2t \, dt$. The expression is $\int \frac{(1 - y)(2 - y)}{y(1 + y)} \frac{dt}{dy} dy$.
Expanding the numerator: $(1 - y)(2 - y) = 2 - 3y + y^2$.
So,$\frac{y^2 - 3y + 2}{y(y + 1)} = 1 + \frac{-4y + 2}{y(y + 1)}$.
Using partial fractions: $\frac{-4y + 2}{y(y + 1)} = \frac{A}{y} + \frac{B}{y + 1} \implies A = 2, B = -6$.
Thus,$I = \int (1 + \frac{2}{t^2} - \frac{6}{1 + t^2}) \, dt = t - \frac{2}{t} - 6 \tan^{-1}(t) + c$.
Substituting $t = \sin x$: $I = \sin x - 2(\sin x)^{-1} - 6 \tan^{-1}(\sin x) + c$.

(B) Given $f(x) = \frac{x^2}{1 + x^3}$.
We need to find $g(x) = \int f(x) \, dx = \int \frac{x^2}{1 + x^3} \, dx$.
Let $u = 1 + x^3$. Then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{du}{3}$.
Substituting these into the integral:
$g(x) = \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \ln|u| + C = \frac{1}{3} \ln(1 + x^3) + C$.
Given the condition $g(1) = 0$:
$0 = \frac{1}{3} \ln(1 + 1^3) + C \implies 0 = \frac{1}{3} \ln(2) + C \implies C = -\frac{1}{3} \ln(2)$.
Substituting $C$ back into the expression for $g(x)$:
$g(x) = \frac{1}{3} \ln(1 + x^3) - \frac{1}{3} \ln(2) = \frac{1}{3} \ln\left( \frac{1 + x^3}{2} \right)$.

(C) Let $I = \int {\frac{{p{x^{p + 2q - 1}} - q{x^{q - 1}}}}{{{{({x^{p + q}} + 1)}^2}}}} \,dx$.
Divide the numerator and denominator by ${x^{2p + 2q}}$:
$I = \int {\frac{{p{x^{p + 2q - 1}} - q{x^{q - 1}}}}{{{x^{2p + 2q}} (1 + {x^{-(p+q)}})^2}}} \,dx$
$I = \int {\frac{{p{x^{-p-1}} - q{x^{-p-q-1}}}}{{{{(1 + {x^{-(p+q)}})}^2}}}} \,dx$.
Let $u = 1 + {x^{-(p+q)}}$.
Then $du = -(p+q) {x^{-(p+q)-1}} \,dx$. This approach is complex.
Alternatively,divide numerator and denominator by ${x^{2q}}$:
$I = \int {\frac{{p{x^{p-1}} - q{x^{-q-1}}}}{{{{({x^p} + {x^{-q}})}^2}}}} \,dx$.
Let $t = {x^p} + {x^{-q}}$.
Then $dt = (p{x^{p-1}} - q{x^{-q-1}}) \,dx$.
Thus,$I = \int {t^{-2}} \,dt = -{t^{-1}} + C = -\frac{1}{{{x^p} + {x^{-q}}}} + C$.
Multiply numerator and denominator by ${x^q}$:
$I = -\frac{{{x^q}}}{{{x^{p+q}} + 1}} + C$.

(C) Let $t = \sec \theta + \tan \theta$.
Then $dt = (\sec \theta \tan \theta + \sec^2 \theta) d\theta = \sec \theta (\tan \theta + \sec \theta) d\theta$.
From $t = \sec \theta + \tan \theta$,we have $\sec \theta - \tan \theta = \frac{1}{t}$.
Adding the two equations: $2 \sec \theta = t + \frac{1}{t} \implies \sec \theta = \frac{t^2 + 1}{2t}$.
Subtracting the two equations: $2 \tan \theta = t - \frac{1}{t} \implies \tan \theta = \frac{t^2 - 1}{2t}$.
Now,$\sec^2 \theta = \sec \theta \cdot \sec \theta = \frac{t^2 + 1}{2t} \cdot \sec \theta$.
Substituting into the integral: $\int \sec^2 \theta (\sec \theta + \tan \theta)^2 d\theta = \int \sec \theta \cdot \sec \theta \cdot t^2 d\theta$.
Since $dt = \sec \theta \cdot t d\theta$,we have $d\theta = \frac{dt}{t \sec \theta}$.
Integral becomes $\int \sec \theta \cdot t^2 \cdot \frac{dt}{t} = \int \sec \theta \cdot t dt = \int \frac{t^2 + 1}{2t} \cdot t dt = \frac{1}{2} \int (t^2 + 1) dt = \frac{1}{2} (\frac{t^3}{3} + t) + C = \frac{t}{6} (t^2 + 3) + C$.
Substituting $t = \sec \theta + \tan \theta$: $\frac{(\sec \theta + \tan \theta)}{6} ((\sec \theta + \tan \theta)^2 + 3) + C$.
Expanding $(\sec \theta + \tan \theta)^2 = \sec^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta = 1 + 2 \tan^2 \theta + 2 \sec \theta \tan \theta$.
Result: $\frac{(\sec \theta + \tan \theta)}{6} (1 + 2 \tan^2 \theta + 2 \sec \theta \tan \theta + 3) = \frac{(\sec \theta + \tan \theta)}{6} (4 + 2 \tan^2 \theta + 2 \sec \theta \tan \theta) = \frac{(\sec \theta + \tan \theta)}{3} (2 + \tan^2 \theta + \sec \theta \tan \theta) = \frac{(\sec \theta + \tan \theta)}{3} [2 + \tan \theta (\tan \theta + \sec \theta)] + C$.

(B) Divide the numerator and denominator by $x^2$:
$\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{\left( {{x^2} + \frac{1}{{{x^2}}} + 3} \right){{\tan }^{ - 1}}\left( {x + \frac{1}{x}} \right)}}} dx$
Since $x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2$,the integral becomes:
$\int {\frac{{(1 - \frac{1}{x^2}) dx}}{{\left( (x + \frac{1}{x})^2 + 1 \right) \tan^{-1}(x + \frac{1}{x})}}}$
Let $u = \tan^{-1}(x + \frac{1}{x})$. Then $du = \frac{1}{1 + (x + \frac{1}{x})^2} \cdot (1 - \frac{1}{x^2}) dx$.
The integral simplifies to $\int \frac{du}{u} = \ln|u| + C$.
Substituting back,we get $\ln|\tan^{-1}(x + \frac{1}{x})| + C$.
Thus,$f(x) = \tan^{-1}(x + \frac{1}{x})$.

(A) Let $I = \int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$.
We can rewrite the argument of the logarithm as $2 + 2x = 2(1 + x)$.
So,$I = \int \frac{\sin(\ln(2(1 + x)))}{x + 1} dx = \int \frac{\sin(\ln 2 + \ln(1 + x))}{x + 1} dx$.
Let $u = \ln(2 + 2x)$.
Then $du = \frac{1}{2 + 2x} \cdot 2 dx = \frac{1}{1 + x} dx$.
Substituting these into the integral,we get $I = \int \sin(u) du$.
The integral of $\sin(u)$ is $-\cos(u) + C$.
Substituting back $u = \ln(2 + 2x)$,we get $I = -\cos(\ln(2 + 2x)) + C$.

(A) We use the substitution $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$.
$\int \frac{dx}{5 + 4\cos x} = \int \frac{dx}{5 + 4\left( \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \right)}$
$= \int \frac{(1 + \tan^2(x/2)) dx}{5(1 + \tan^2(x/2)) + 4(1 - \tan^2(x/2))}$
$= \int \frac{\sec^2(x/2) dx}{9 + \tan^2(x/2)}$
Let $\tan(x/2) = y$,then $\frac{1}{2} \sec^2(x/2) dx = dy$,so $\sec^2(x/2) dx = 2 dy$.
$= \int \frac{2 dy}{9 + y^2} = 2 \int \frac{dy}{3^2 + y^2} = 2 \cdot \frac{1}{3} \tan^{-1} \left( \frac{y}{3} \right) + C$
$= \frac{2}{3} \tan^{-1} \left( \frac{1}{3} \tan \frac{x}{2} \right) + C$.
Comparing with $\lambda \tan^{-1} (m \tan(x/2))$,we get $\lambda = 2/3$ and $m = 1/3$.

(D) Let $I = \int \frac{1}{x^2 - 1} \ln \left( \frac{x - 1}{x + 1} \right) dx$.
We know that $\frac{1}{x^2 - 1} = \frac{1}{(x-1)(x+1)}$.
Let $t = \ln \left( \frac{x - 1}{x + 1} \right) = \ln(x-1) - \ln(x+1)$.
Differentiating with respect to $x$:
$dt = \left( \frac{1}{x-1} - \frac{1}{x+1} \right) dx = \left( \frac{(x+1) - (x-1)}{(x-1)(x+1)} \right) dx = \frac{2}{x^2 - 1} dx$.
Thus,$\frac{1}{x^2 - 1} dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \int t \cdot \frac{1}{2} dt = \frac{1}{2} \cdot \frac{t^2}{2} + C = \frac{1}{4} t^2 + C$.
Substituting $t$ back:
$I = \frac{1}{4} \ln^2 \left( \frac{x - 1}{x + 1} \right) + C$.
Since $\ln \left( \frac{x + 1}{x - 1} \right) = - \ln \left( \frac{x - 1}{x + 1} \right)$,squaring it gives the same result:
$\ln^2 \left( \frac{x + 1}{x - 1} \right) = \ln^2 \left( \frac{x - 1}{x + 1} \right)$.
Therefore,both $(B)$ and $(C)$ are correct.

(B) Let $I = \int \frac{(2x + 3)dx}{x(x + 1)(x + 2)(x + 3) + 1}$.
Rearrange the denominator: $[x(x + 3)][(x + 1)(x + 2)] + 1 = (x^2 + 3x)(x^2 + 3x + 2) + 1$.
Let $t = x^2 + 3x$. Then $dt = (2x + 3)dx$.
The integral becomes $I = \int \frac{dt}{t(t + 2) + 1} = \int \frac{dt}{t^2 + 2t + 1} = \int \frac{dt}{(t + 1)^2}$.
Integrating,we get $I = -\frac{1}{t + 1} + C = C - \frac{1}{x^2 + 3x + 1}$.
Comparing this with $C - \frac{1}{f(x)}$,we have $f(x) = x^2 + 3x + 1$.
Thus,$a = 1, b = 3, c = 1$.
Therefore,$a + b + c = 1 + 3 + 1 = 5$.

(A) Given integral is $I = \int \frac{\sec x(1 + \tan x)dx}{e^{-x} + \sec x}$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{e^x \sec x(1 + \tan x)dx}{1 + e^x \sec x}$.
Let $t = 1 + e^x \sec x$. Then $dt = (e^x \sec x + e^x \sec x \tan x) dx = e^x \sec x(1 + \tan x) dx$.
Thus,$I = \int \frac{dt}{t} = \ln |t| + C = \ln |1 + e^x \sec x| + C$.
So,$f(x) = \ln(1 + e^x \sec x) + C$.
Given $f(0) = \ln 2$,we have $\ln(1 + e^0 \sec 0) + C = \ln 2 \Rightarrow \ln(1 + 1) + C = \ln 2 \Rightarrow \ln 2 + C = \ln 2 \Rightarrow C = 0$.
Therefore,$f(x) = \ln(1 + e^x \sec x)$.
Now,$f\left( \frac{\pi}{4} \right) = \ln(1 + e^{\frac{\pi}{4}} \sec \frac{\pi}{4}) = \ln(1 + e^{\frac{\pi}{4}} \sqrt{2})$.
The correct option is $A$.

(D) Let $\sin \phi = t$,then $\cos \phi \, d\phi = dt$.
Substituting this into the integral:
$I = \int \frac{(3t - 2) dt}{5 - (1 - t^2) - 4t} = \int \frac{3t - 2}{t^2 - 4t + 4} dt = \int \frac{3t - 2}{(t - 2)^2} dt$.
Using partial fractions:
$\frac{3t - 2}{(t - 2)^2} = \frac{A}{t - 2} + \frac{B}{(t - 2)^2}$.
$3t - 2 = A(t - 2) + B$.
Comparing coefficients,$A = 3$ and $B = 4$.
Thus,$I = \int \frac{3}{t - 2} dt + \int \frac{4}{(t - 2)^2} dt$.
$I = 3 \log |t - 2| - \frac{4}{t - 2} + C$.
Since $t = \sin \phi$,we have $I = 3 \log |\sin \phi - 2| - \frac{4}{\sin \phi - 2} + C$.
This simplifies to $I = 3 \log (2 - \sin \phi) + \frac{4}{2 - \sin \phi} + C$.

(D) Given $f(x) = \frac{x}{(1 + x^7)^{1/7}}$.
Calculating $f(f(x)) = \frac{f(x)}{(1 + f(x)^7)^{1/7}} = \frac{x/(1+x^7)^{1/7}}{(1 + x^7/(1+x^7))^{1/7}} = \frac{x/(1+x^7)^{1/7}}{(1/(1+x^7))^{1/7}} = \frac{x}{(1+2x^7)^{1/7}}$.
By induction,$g(x) = (f \circ f \circ f \circ f \circ f \circ f \circ f)(x) = \frac{x}{(1 + 7x^7)^{1/7}}$.
We need to evaluate $I = \int x^5 g(x) dx = \int \frac{x^6}{(1 + 7x^7)^{1/7}} dx$.
Let $u = 1 + 7x^7$,then $du = 49x^6 dx$,so $x^6 dx = \frac{du}{49}$.
$I = \int \frac{1}{u^{1/7}} \cdot \frac{du}{49} = \frac{1}{49} \int u^{-1/7} du = \frac{1}{49} \cdot \frac{u^{6/7}}{6/7} + C = \frac{1}{49} \cdot \frac{7}{6} u^{6/7} + C = \frac{1}{42} (1 + 7x^7)^{6/7} + C$.

(B) Given integral is $I = \int \frac{x^{5m-1} + 2x^{4m-1}}{(x^{2m} + x^m + 1)^3} dx$.
Divide numerator and denominator by $x^{6m}$:
$I = \int \frac{x^{-m-1} + 2x^{-2m-1}}{(1 + x^{-m} + x^{-2m})^3} dx$.
Let $t = 1 + x^{-m} + x^{-2m}$.
Then $dt = (-mx^{-m-1} - 2mx^{-2m-1}) dx = -m(x^{-m-1} + 2x^{-2m-1}) dx$.
So,$(x^{-m-1} + 2x^{-2m-1}) dx = -\frac{1}{m} dt$.
Substituting these into the integral:
$I = \int \frac{-1/m}{t^3} dt = -\frac{1}{m} \int t^{-3} dt = -\frac{1}{m} \left( \frac{t^{-2}}{-2} \right) + c = \frac{1}{2mt^2} + c$.
Substituting $t = 1 + x^{-m} + x^{-2m} = \frac{x^{2m} + x^m + 1}{x^{2m}}$:
$f(x) = \frac{1}{2m \left( \frac{x^{2m} + x^m + 1}{x^{2m}} \right)^2} = \frac{x^{4m}}{2m(x^{2m} + x^m + 1)^2} + c$.

(C) Let $I = \int \frac{x^{1/3}}{(1 + x^{2/3})^3} dx$.
Substitute $u = 1 + x^{2/3}$.
Then $du = \frac{2}{3} x^{-1/3} dx$,which implies $dx = \frac{3}{2} x^{1/3} du$.
Substituting this into the integral:
$I = \int \frac{x^{1/3}}{u^3} \cdot \frac{3}{2} x^{1/3} du = \frac{3}{2} \int \frac{x^{2/3}}{u^3} du$.
Since $x^{2/3} = u - 1$,we have:
$I = \frac{3}{2} \int \frac{u - 1}{u^3} du = \frac{3}{2} \int (u^{-2} - u^{-3}) du$.
Integrating term by term:
$I = \frac{3}{2} \left( \frac{u^{-1}}{-1} - \frac{u^{-2}}{-2} \right) + C = \frac{3}{2} \left( -\frac{1}{u} + \frac{1}{2u^2} \right) + C = \frac{3}{2} \left( \frac{-2u + 1}{2u^2} \right) + C$.
Substituting $u = 1 + x^{2/3}$ back:
$I = \frac{3}{4} \left( \frac{-2(1 + x^{2/3}) + 1}{(1 + x^{2/3})^2} \right) + C = \frac{3}{4} \left( \frac{-2 - 2x^{2/3} + 1}{(1 + x^{2/3})^2} \right) + C$.
Wait,let's re-evaluate the substitution method:
$I = \int \frac{x^{1/3}}{(1 + x^{2/3})^3} dx$.
Let $x^{2/3} = t$,then $\frac{2}{3} x^{-1/3} dx = dt \implies dx = \frac{3}{2} x^{1/3} dt = \frac{3}{2} t^{1/2} dt$.
$I = \int \frac{t^{1/2}}{(1 + t)^3} \cdot \frac{3}{2} t^{1/2} dt = \frac{3}{2} \int \frac{t}{(1 + t)^3} dt$.
$I = \frac{3}{2} \int \frac{t + 1 - 1}{(1 + t)^3} dt = \frac{3}{2} \int ((1 + t)^{-2} - (1 + t)^{-3}) dt$.
$I = \frac{3}{2} \left( \frac{(1 + t)^{-1}}{-1} - \frac{(1 + t)^{-2}}{-2} \right) + C = \frac{3}{2} \left( -\frac{1}{1 + t} + \frac{1}{2(1 + t)^2} \right) + C$.
$I = \frac{3}{2} \left( \frac{-2(1 + t) + 1}{2(1 + t)^2} \right) + C = \frac{3}{4} \left( \frac{-2 - 2t + 1}{(1 + t)^2} \right) + C = \frac{3}{4} \left( \frac{-1 - 2t}{(1 + t)^2} \right) + C$.
Given the options,the intended form is $\frac{3}{4} \frac{x^{4/3}}{(1+x^{2/3})^2} + C$.

(A) Given $I = \int \frac{\sin^2 x - 1}{2x \sin^2 x + \sin 2x} dx$.
Since $\sin^2 x - 1 = -\cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:
$I = \int \frac{-\cos^2 x}{2x \sin^2 x + 2 \sin x \cos x} dx = \int \frac{-\cos^2 x}{2 \sin x (x \sin x + \cos x)} dx$.
Divide numerator and denominator by $\sin^2 x$:
$I = \int \frac{-\cot^2 x}{2(x + \cot x)} dx$.
Let $u = x + \cot x$,then $du = (1 - \csc^2 x) dx = (1 - (1 + \cot^2 x)) dx = -\cot^2 x dx$.
Substituting these into the integral:
$I = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + c = \frac{1}{2} \ln|x + \cot x| + c$.
Thus,$2I = \ln|x + \cot x| + 2c$.
Taking the exponential of both sides:
$e^{2I} = e^{\ln|x + \cot x| + 2c} = |x + \cot x| e^{2c}$.
Let $C = e^{2c}$,then $Ce^{2I} = x + \cot x = x + \frac{\cos x}{\sin x} = \frac{x \sin x + \cos x}{\sin x}$.
Therefore,$Ce^{2I} \sin x = x \sin x + \cos x$.

(D) Let $I = \int \frac{xdx}{\sqrt{1 + x^2 + (1 + x^2)^{3/2}}}$.
Substitute $1 + x^2 = t^2$,then $2xdx = 2tdt$,which implies $xdx = tdt$.
Substituting these into the integral:
$I = \int \frac{tdt}{\sqrt{t^2 + (t^2)^{3/2}}} = \int \frac{tdt}{\sqrt{t^2 + t^3}}$.
Since $t > 0$,$\sqrt{t^2 + t^3} = t\sqrt{1 + t}$.
$I = \int \frac{tdt}{t\sqrt{1 + t}} = \int \frac{dt}{\sqrt{1 + t}}$.
Integrating with respect to $t$:
$I = 2\sqrt{1 + t} + C$.
Substituting back $t = \sqrt{1 + x^2}$:
$I = 2\sqrt{1 + \sqrt{1 + x^2}} + C$.

(A) Let $I = \int \frac{2x^2}{x^3+1} dx$.
Substitute $u = x^3+1$,then $du = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} du$.
Substituting these into the integral:
$I = \int \frac{2}{u} \cdot \frac{1}{3} du = \frac{2}{3} \int \frac{1}{u} du$.
$I = \frac{2}{3} \ln |u| + C$.
Substituting $u = x^3+1$ back,we get $I = \frac{2}{3} \ln |x^3+1| + C$.

(A) Given integral: $I = \int (6x^2 + 5x + 4)(x^2 + x + 1)^6 \cdot x^{27} dx$
We can rewrite the expression as:
$I = \int (x^2 + x + 1)^6 \cdot x^{24} \cdot (6x^2 + 5x + 4)x^3 dx$
$I = \int (x^2 + x + 1)^6 \cdot x^{24} \cdot (6x^5 + 5x^4 + 4x^3) dx$
$I = \int (x^2(x^2 + x + 1))^6 \cdot (6x^5 + 5x^4 + 4x^3) dx$
$I = \int (x^6 + x^5 + x^4)^6 \cdot (6x^5 + 5x^4 + 4x^3) dx$
Let $t = x^6 + x^5 + x^4$.
Then $dt = (6x^5 + 5x^4 + 4x^3) dx$.
Substituting these into the integral:
$I = \int t^6 dt = \frac{t^7}{7} + C$
$I = \frac{(x^6 + x^5 + x^4)^7}{7} + C$
$I = \frac{(x^4(x^2 + x + 1))^7}{7} + C$
$I = \frac{x^{28}(x^2 + x + 1)^7}{7} + C$

(C) Let $I = \int \frac{1 - {x^7}}{x(1 + {x^7})} dx$.
We can rewrite the integrand as:
$\frac{1 - {x^7}}{x(1 + {x^7})} = \frac{(1 + {x^7}) - 2{x^7}}{x(1 + {x^7})} = \frac{1}{x} - \frac{2{x^7}}{x(1 + {x^7})} = \frac{1}{x} - \frac{2{x^6}}{1 + {x^7}}$.
Now,integrate term by term:
$I = \int \frac{1}{x} dx - \int \frac{2{x^6}}{1 + {x^7}} dx$.
For the second integral,let $u = 1 + {x^7}$,then $du = 7{x^6} dx$,which implies ${x^6} dx = \frac{du}{7}$.
Substituting these into the integral:
$I = \ln |x| - 2 \int \frac{1}{u} \cdot \frac{du}{7} = \ln |x| - \frac{2}{7} \ln |u| + c$.
Substituting $u = 1 + {x^7}$ back:
$I = \ln |x| - \frac{2}{7} \ln (1 + {x^7}) + c$.

(D) Let $I = \int \frac{x^7 + x^2 + 1}{(3x^8 + 8x^3 + 24x)^{1/3}} dx$.
Consider the substitution $t = 3x^8 + 8x^3 + 24x$.
Then,differentiating with respect to $x$,we get $dt = (24x^7 + 24x^2 + 24) dx$.
This simplifies to $dt = 24(x^7 + x^2 + 1) dx$,which implies $(x^7 + x^2 + 1) dx = \frac{1}{24} dt$.
Substituting these into the integral,we have $I = \int \frac{1}{t^{1/3}} \cdot \frac{1}{24} dt$.
$I = \frac{1}{24} \int t^{-1/3} dt = \frac{1}{24} \cdot \frac{t^{2/3}}{2/3} + C$.
$I = \frac{1}{24} \cdot \frac{3}{2} t^{2/3} + C = \frac{1}{16} t^{2/3} + C$.
Substituting back $t = 3x^8 + 8x^3 + 24x$,we get $I = \frac{1}{16} (3x^8 + 8x^3 + 24x)^{2/3} + C$.

(A) Let $I = \int \frac{dx}{x^2(1+x^4)^{3/4}}$.
We can rewrite the integral by taking $x^4$ common from the bracket:
$I = \int \frac{dx}{x^2 \cdot (x^4(x^{-4}+1))^{3/4}} = \int \frac{dx}{x^2 \cdot x^3(1+x^{-4})^{3/4}} = \int \frac{dx}{x^5(1+x^{-4})^{3/4}}$.
Let $t = 1+x^{-4}$.
Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{1}{4} dt}{t^{3/4}} = -\frac{1}{4} \int t^{-3/4} dt$.
Integrating with respect to $t$:
$I = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + C = -t^{1/4} + C$.
Substituting $t = 1+x^{-4}$ back:
$I = -(1+x^{-4})^{1/4} + C = -(\frac{x^4+1}{x^4})^{1/4} + C = -\frac{(1+x^4)^{1/4}}{x} + C$.

(C) Let $I = \int \sin^{\frac{-1}{2}}x \cos^{\frac{-7}{2}}x \, dx$.
We can rewrite the integrand as:
$I = \int \frac{1}{\sin^{\frac{1}{2}}x \cos^{\frac{7}{2}}x} \, dx = \int \frac{1}{\left(\frac{\sin x}{\cos x}\right)^{\frac{1}{2}} \cos^4 x} \, dx$.
$I = \int \frac{\sec^4 x}{\tan^{\frac{1}{2}}x} \, dx = \int \frac{\sec^2 x \cdot \sec^2 x}{\tan^{\frac{1}{2}}x} \, dx$.
Since $\sec^2 x = 1 + \tan^2 x$,we have:
$I = \int \frac{(1 + \tan^2 x) \sec^2 x}{\tan^{\frac{1}{2}}x} \, dx$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
$I = \int \frac{1 + u^2}{u^{\frac{1}{2}}} \, du = \int (u^{-\frac{1}{2}} + u^{\frac{3}{2}}) \, du$.
Integrating with respect to $u$:
$I = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + C = 2u^{\frac{1}{2}} + \frac{2}{5}u^{\frac{5}{2}} + C$.
Substituting $u = \tan x$ back:
$I = 2\tan^{\frac{1}{2}}x + \frac{2}{5}\tan^{\frac{5}{2}}x + C$.

(C) Let $I = \int \frac{p x^{p+2q-1} - q x^{q-1}}{(x^{p+q} + 1)^2} dx$.
Divide the numerator and denominator by $x^{2p+2q}$ or manipulate the expression:
$I = \int \frac{p x^{p-1} - q x^{-q-1}}{(x^p + x^{-q})^2} dx$.
Let $t = x^p + x^{-q}$.
Then $dt = (p x^{p-1} - q x^{-q-1}) dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t^2} dt = -\frac{1}{t} + C$.
Substituting back $t = x^p + x^{-q} = \frac{x^{p+q} + 1}{x^q}$:
$I = -\frac{1}{\frac{x^{p+q} + 1}{x^q}} + C = -\frac{x^q}{x^{p+q} + 1} + C$.

(C) Let $u = x^2 + 2$. Then $du = 2x \, dx$.
We want to find the integral $\int (x^2 + 4)^{-1/2} \, d(x^2 + 2)$.
Since $d(x^2 + 2) = 2x \, dx$,the integral becomes $\int \frac{1}{\sqrt{x^2 + 4}} \cdot 2x \, dx$.
Let $t = x^2 + 4$. Then $dt = 2x \, dx$.
The integral becomes $\int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt$.
Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$,we get:
$\frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$.
Substituting back $t = x^2 + 4$,we get $2\sqrt{x^2 + 4} + C$.

(A) Let $I = \int \frac{dx}{x(x^4 - 1)}$.
Multiply the numerator and denominator by $x^3$:
$I = \int \frac{x^3 dx}{x^4(x^4 - 1)}$.
Let $u = x^4$,then $du = 4x^3 dx$,which implies $x^3 dx = \frac{du}{4}$.
Substituting these into the integral:
$I = \frac{1}{4} \int \frac{du}{u(u - 1)}$.
Using partial fractions:
$\frac{1}{u(u - 1)} = \frac{1}{u - 1} - \frac{1}{u}$.
Therefore,$I = \frac{1}{4} \int \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$.
$I = \frac{1}{4} (\ln |u - 1| - \ln |u|) + C$.
$I = \frac{1}{4} \ln \left| \frac{u - 1}{u} \right| + C$.
Substituting $u = x^4$ back:
$I = \frac{1}{4} \ln \left| \frac{x^4 - 1}{x^4} \right| + C$.

(C) Let $I = \int \frac{\sec^2 x}{\log ((\tan x)^{\tan x})} dx$.
Using the property $\log(a^b) = b \log a$,the denominator becomes $\tan x \cdot \log(\tan x)$.
So,$I = \int \frac{\sec^2 x}{\tan x \cdot \log(\tan x)} dx$.
Let $\tan x = t$,then $\sec^2 x dx = dt$.
Substituting these into the integral,we get $I = \int \frac{dt}{t \log t}$.
Now,let $\log t = z$,then $\frac{1}{t} dt = dz$.
Substituting these,we get $I = \int \frac{dz}{z} = \log |z| + c$.
Replacing $z$ with $\log t$,we get $I = \log |\log t| + c$.
Finally,replacing $t$ with $\tan x$,we get $I = \log |\log(\tan x)| + c$.

(D) Let $t = \tan^{-1} \sqrt{x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = \frac{1}{1 + (\sqrt{x})^2} \times \frac{1}{2\sqrt{x}} = \frac{1}{1+x} \times \frac{1}{2\sqrt{x}}$.
Thus,$dt = \frac{dx}{2\sqrt{x}(1+x)} = \frac{dx}{2(\sqrt{x} + x\sqrt{x})}$.
This implies $\frac{dx}{\sqrt{x} + x\sqrt{x}} = 2 dt$.
Substituting these into the integral:
$\int e^t (2 dt) = 2 \int e^t dt = 2e^t + c$.
Replacing $t$ with $\tan^{-1} \sqrt{x}$,we get:
$2e^{\tan^{-1} \sqrt{x}} + c$.

(B) Let $I = \int \frac{\cos x + x \sin x}{x(x - \cos x)} dx$.
We can rewrite the denominator as $x^2(1 - \frac{\cos x}{x})$.
So,$I = \int \frac{\cos x + x \sin x}{x^2(1 - \frac{\cos x}{x})} dx$.
Let $t = 1 - \frac{\cos x}{x}$.
Then,differentiating with respect to $x$ using the quotient rule:
$dt = - \left( \frac{x(-\sin x) - \cos x(1)}{x^2} \right) dx = \frac{x \sin x + \cos x}{x^2} dx$.
Substituting these into the integral:
$I = \int \frac{1}{t} dt = \ln |t| + c$.
Substituting back $t = 1 - \frac{\cos x}{x}$:
$I = \ln \left| 1 - \frac{\cos x}{x} \right| + c$.

(B) Let $I = \int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$.
Rewrite the integrand: $\sqrt{\tan x} + \sqrt{\cot x} = \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} = \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}$.
Multiply numerator and denominator by $\sqrt{2}$: $\frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} = \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{\sin 2x}}$.
Since $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 1 - (\sin x - \cos x)^2$,we have:
$I = \sqrt{2} \int \frac{(\sin x + \cos x) \, dx}{\sqrt{1 - (\sin x - \cos x)^2}}$.
Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) \, dx$.
Substituting this into the integral: $I = \sqrt{2} \int \frac{du}{\sqrt{1 - u^2}} = \sqrt{2} \sin^{-1}(u) + C$.
Thus,$I = \sqrt{2} \sin^{-1}(\sin x - \cos x) + C$.

(A) Let $I = \int \frac{dx}{(x+1)^{3/4}(x-2)^{5/4}}$.
We can rewrite the integral as:
$I = \int \frac{dx}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^{3/4} (x-2)^{5/4}} = \int \frac{dx}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^2}$.
Let $t = \frac{x+1}{x-2}$.
Then,$dt = \frac{(x-2)(1) - (x+1)(1)}{(x-2)^2} dx = \frac{x-2-x-1}{(x-2)^2} dx = \frac{-3}{(x-2)^2} dx$.
So,$\frac{dx}{(x-2)^2} = -\frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \frac{1}{t^{3/4}} \left(-\frac{1}{3} dt\right) = -\frac{1}{3} \int t^{-3/4} dt$.
Integrating with respect to $t$:
$I = -\frac{1}{3} \left( \frac{t^{-3/4 + 1}}{-3/4 + 1} \right) + c = -\frac{1}{3} \left( \frac{t^{1/4}}{1/4} \right) + c = -\frac{4}{3} t^{1/4} + c$.
Substituting $t = \frac{x+1}{x-2}$ back:
$I = -\frac{4}{3} \left( \frac{x+1}{x-2} \right)^{1/4} + c$.

(D) Let $I = \int \frac{\log (t+\sqrt{1+t^{2}})}{\sqrt{1+t^{2}}} dt$.
Substitute $u = \log (t+\sqrt{1+t^{2}})$.
Then,$du = \frac{1}{t+\sqrt{1+t^{2}}} \cdot \left( 1 + \frac{2t}{2\sqrt{1+t^{2}}} \right) dt = \frac{1}{t+\sqrt{1+t^{2}}} \cdot \left( \frac{\sqrt{1+t^{2}}+t}{\sqrt{1+t^{2}}} \right) dt = \frac{1}{\sqrt{1+t^{2}}} dt$.
Therefore,$I = \int u du = \frac{u^{2}}{2} + C$.
Comparing this with the given expression $I = \frac{1}{2}[g(t)]^{2} + C$,we get $g(t) = u = \log (t+\sqrt{1+t^{2}})$.
Now,evaluating at $t = 2$:
$g(2) = \log (2+\sqrt{1+2^{2}}) = \log (2+\sqrt{5})$.

(B) Let $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$
Divide numerator and denominator by $\cos^6 x$:
$I = \int \frac{\frac{\sin^2 x \cos^2 x}{\cos^6 x}}{(\frac{\sin^3 x + \cos^3 x}{\cos^3 x})^2} dx$
$I = \int \frac{\tan^2 x \sec^2 x}{(1 + \tan^3 x)^2} dx$
Let $1 + \tan^3 x = t$. Then $3 \tan^2 x \sec^2 x dx = dt$,so $\tan^2 x \sec^2 x dx = \frac{dt}{3}$.
Substituting these into the integral:
$I = \int \frac{1}{t^2} \cdot \frac{dt}{3} = \frac{1}{3} \int t^{-2} dt$
$I = \frac{1}{3} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{3t} + c$
Substituting back $t = 1 + \tan^3 x$:
$I = -\frac{1}{3(1 + \tan^3 x)} + c$

(B) Let $I = \int \frac{x dx}{2 - x^2 + \sqrt{2 - x^2}}$.
Substitute $t = \sqrt{2 - x^2}$. Then $t^2 = 2 - x^2$,which implies $2t dt = -2x dx$,or $x dx = -t dt$.
Substituting these into the integral:
$I = \int \frac{-t dt}{t^2 + t}$
Simplify the integrand:
$I = \int \frac{-t dt}{t(t + 1)} = -\int \frac{dt}{t + 1}$
Integrating with respect to $t$:
$I = -\log |t + 1| + c$
Substituting back $t = \sqrt{2 - x^2}$:
$I = -\log |\sqrt{2 - x^2} + 1| + c$.

(A) We have $f(x) = \int \csc^6 x \, dx$.
Using the identity $\csc^2 x = 1 + \cot^2 x$,we can write:
$f(x) = \int \csc^4 x \cdot \csc^2 x \, dx = \int (1 + \cot^2 x)^2 \csc^2 x \, dx$.
Let $u = \cot x$,then $du = -\csc^2 x \, dx$,which implies $\csc^2 x \, dx = -du$.
Substituting these into the integral:
$f(x) = \int (1 + u^2)^2 (-du) = -\int (1 + 2u^2 + u^4) \, du$.
Integrating with respect to $u$:
$f(x) = -(u + \frac{2u^3}{3} + \frac{u^5}{5}) + C$.
Substituting $u = \cot x$ back:
$f(x) = -\cot x - \frac{2}{3} \cot^3 x - \frac{1}{5} \cot^5 x + C$.
This is a polynomial in $\cot x$ of degree $5$.

(D) Let $I = \int x \sqrt{\frac{2 \sin(x^2-1) - 2 \sin(x^2-1) \cos(x^2-1)}{2 \sin(x^2-1) + 2 \sin(x^2-1) \cos(x^2-1)}} dx$
$I = \int x \sqrt{\frac{2 \sin(x^2-1) [1 - \cos(x^2-1)]}{2 \sin(x^2-1) [1 + \cos(x^2-1)]}} dx$
$I = \int x \sqrt{\frac{2 \sin^2(\frac{x^2-1}{2})}{2 \cos^2(\frac{x^2-1}{2})}} dx$
$I = \int x \tan(\frac{x^2-1}{2}) dx$
Let $t = \frac{x^2-1}{2}$,then $dt = x dx$
$I = \int \tan(t) dt = \ln|\sec(t)| + c$
$I = \ln|\sec(\frac{x^2-1}{2})| + c$

(A) Let $I = \int \frac{(\sin^n \theta - \sin \theta)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d\theta$.
Substitute $\sin \theta = t$,then $\cos \theta d\theta = dt$.
The integral becomes $I = \int \frac{(t^n - t)^{\frac{1}{n}}}{t^{n+1}} dt$.
Factor out $t^n$ from the numerator: $I = \int \frac{(t^n(1 - t^{1-n}))^{\frac{1}{n}}}{t^{n+1}} dt = \int \frac{t(1 - t^{1-n})^{\frac{1}{n}}}{t^{n+1}} dt = \int \frac{(1 - t^{1-n})^{\frac{1}{n}}}{t^n} dt$.
Let $z = 1 - t^{1-n}$. Then $dz = -(1-n)t^{-n} dt = (n-1)t^{-n} dt$.
Thus,$t^{-n} dt = \frac{dz}{n-1}$.
Substituting into the integral: $I = \frac{1}{n-1} \int z^{\frac{1}{n}} dz$.
Integrating with respect to $z$: $I = \frac{1}{n-1} \cdot \frac{z^{\frac{1}{n} + 1}}{\frac{1}{n} + 1} + C = \frac{1}{n-1} \cdot \frac{z^{\frac{n+1}{n}}}{\frac{n+1}{n}} + C = \frac{n}{(n-1)(n+1)} z^{\frac{n+1}{n}} + C$.
$I = \frac{n}{n^2 - 1} (1 - t^{1-n})^{\frac{n+1}{n}} + C$.
Substituting $t = \sin \theta$ back: $I = \frac{n}{n^2 - 1} (1 - \frac{1}{\sin^{n-1} \theta})^{\frac{n+1}{n}} + C$.

(D) Let $2x - 1 = t^2$. Then $2 dx = 2t dt$,so $dx = t dt$.
Also,$x = \frac{t^2 + 1}{2}$,which implies $x + 1 = \frac{t^2 + 1}{2} + 1 = \frac{t^2 + 3}{2}$.
Substituting these into the integral:
$\int \frac{x + 1}{\sqrt{2x - 1}} dx = \int \frac{(\frac{t^2 + 3}{2})}{t} (t dt) = \int \frac{t^2 + 3}{2} dt$
$= \frac{1}{2} (\frac{t^3}{3} + 3t) + C = \frac{t^3}{6} + \frac{3t}{2} + C$
$= t (\frac{t^2}{6} + \frac{3}{2}) + C = t (\frac{t^2 + 9}{6}) + C$
Since $t = \sqrt{2x - 1}$,we have $t^2 = 2x - 1$.
Substituting $t^2$ back:
$= \sqrt{2x - 1} (\frac{2x - 1 + 9}{6}) + C = \sqrt{2x - 1} (\frac{2x + 8}{6}) + C$
$= \sqrt{2x - 1} (\frac{x + 4}{3}) + C$.
Comparing this with $f(x) \sqrt{2x - 1} + C$,we get $f(x) = \frac{1}{3}(x + 4)$.

(D) Let $I = \int \sec^{2/3} x \csc^{4/3} x \, dx$.
We can rewrite the integrand as:
$I = \int \frac{1}{(\cos x)^{2/3} (\sin x)^{4/3}} \, dx$.
Divide the numerator and denominator by $\cos^{4/3} x$:
$I = \int \frac{1}{\frac{(\sin x)^{4/3}}{(\cos x)^{4/3}} \cdot (\cos x)^{2/3} \cdot (\cos x)^{4/3}} \, dx$.
$I = \int \frac{1}{(\tan x)^{4/3} \cdot \cos^2 x} \, dx$.
$I = \int \frac{\sec^2 x}{(\tan x)^{4/3}} \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$.
Substituting these into the integral:
$I = \int t^{-4/3} \, dt$.
Integrating with respect to $t$:
$I = \frac{t^{-4/3 + 1}}{-4/3 + 1} + C = \frac{t^{-1/3}}{-1/3} + C = -3 t^{-1/3} + C$.
Substituting back $t = \tan x$:
$I = -3 \tan^{-1/3} x + C$.

(C) Let $I = \int \frac{2x^3 - 1}{x^4 + x} \,dx$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{2x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} \,dx$.
Let $t = x^2 + \frac{1}{x}$.
Then $dt = (2x - \frac{1}{x^2}) \,dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t} \,dt = \log_e |t| + C$.
Substituting back $t = x^2 + \frac{1}{x} = \frac{x^3 + 1}{x}$:
$I = \log_e \left| \frac{x^3 + 1}{x} \right| + C$.

(A) Let $I = \int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}$.
We can rewrite the integrand as $I = \int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{\frac{8}{7}}(x-3)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}} = \int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{\frac{8}{7}}(x-3)^{2}}$.
Let $t = \frac{x+4}{x-3}$. Then $dt = \frac{(x-3)(1) - (x+4)(1)}{(x-3)^2} dx = \frac{-7}{(x-3)^2} dx$,which implies $\frac{dx}{(x-3)^2} = -\frac{1}{7} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{t^{\frac{8}{7}}} \left(-\frac{1}{7} dt\right) = -\frac{1}{7} \int t^{-\frac{8}{7}} dt$.
Integrating,we get $I = -\frac{1}{7} \left( \frac{t^{-\frac{8}{7} + 1}}{-\frac{8}{7} + 1} \right) + C = -\frac{1}{7} \left( \frac{t^{-\frac{1}{7}}}{-\frac{1}{7}} \right) + C = t^{-\frac{1}{7}} + C$.
Substituting $t = \frac{x+4}{x-3}$ back,we get $I = \left(\frac{x+4}{x-3}\right)^{-\frac{1}{7}} + C = \left(\frac{x-3}{x+4}\right)^{\frac{1}{7}} + C$.

(A) To integrate $\int 2 x \sin \left(x^{2}+1\right) d x$,we use the method of substitution.
Let $t = x^{2}+1$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = 2x$,which implies $dt = 2x \, dx$.
Substituting these into the integral,we get:
$\int \sin(t) \, dt = -\cos(t) + C$.
Finally,substituting $t = x^{2}+1$ back into the expression,we obtain:
$-\cos \left(x^{2}+1\right) + C$.

(N/A) Let $I = \int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} dx$.
Step $1$: Use substitution $\sqrt{x} = t$. Then $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$.
Step $2$: Substitute into the integral:
$I = \int \tan^4 t \cdot \sec^2 t \cdot (2 dt) = 2 \int \tan^4 t \sec^2 t dt$.
Step $3$: Use another substitution $\tan t = u$. Then $\sec^2 t dt = du$.
Step $4$: Integrate with respect to $u$:
$I = 2 \int u^4 du = 2 \cdot \frac{u^5}{5} + C = \frac{2}{5} u^5 + C$.
Step $5$: Substitute back $u = \tan t$ and $t = \sqrt{x}$:
$I = \frac{2}{5} \tan^5 \sqrt{x} + C$.

(A) Let $I = \int \frac{\sin(\tan^{-1} x)}{1+x^2} dx$.
We know that the derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$.
Let $t = \tan^{-1} x$.
Then,$dt = \frac{1}{1+x^2} dx$.
Substituting these into the integral,we get:
$I = \int \sin(t) dt$.
Integrating $\sin(t)$ with respect to $t$,we get:
$I = -\cos(t) + C$.
Substituting back $t = \tan^{-1} x$,we get:
$I = -\cos(\tan^{-1} x) + C$.

(A) We have $\int \sin ^{3} x \cos ^{2} x \, dx = \int \sin ^{2} x \cos ^{2} x (\sin x) \, dx$
$= \int (1 - \cos ^{2} x) \cos ^{2} x (\sin x) \, dx$
Let $t = \cos x$,then $dt = -\sin x \, dx$,which implies $\sin x \, dx = -dt$
Substituting these into the integral,we get:
$= -\int (1 - t^{2}) t^{2} \, dt$
$= -\int (t^{2} - t^{4}) \, dt$
$= -(\frac{t^{3}}{3} - \frac{t^{5}}{5}) + C$
$= -\frac{1}{3} t^{3} + \frac{1}{5} t^{5} + C$
Substituting $t = \cos x$ back,we get:
$= -\frac{1}{3} \cos ^{3} x + \frac{1}{5} \cos ^{5} x + C$