Definite integration by substitution Questions in English

(A) Given the integral $\int_\alpha^{\log _e 4} \frac{dx}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$.
Let $e^{x}-1=t^2$,then $e^x dx = 2t dt$,which implies $dx = \frac{2t dt}{t^2+1}$.
When $x = \log_e 4$,$t = \sqrt{e^{\log_e 4}-1} = \sqrt{4-1} = \sqrt{3}$.
When $x = \alpha$,$t = \sqrt{e^\alpha-1}$.
The integral becomes $\int_{\sqrt{e^\alpha-1}}^{\sqrt{3}} \frac{2t dt}{t(t^2+1)} = 2 \int_{\sqrt{e^\alpha-1}}^{\sqrt{3}} \frac{dt}{t^2+1} = 2 [\tan^{-1} t]_{\sqrt{e^\alpha-1}}^{\sqrt{3}}$.
This equals $2(\tan^{-1} \sqrt{3} - \tan^{-1} \sqrt{e^\alpha-1}) = 2(\frac{\pi}{3} - \tan^{-1} \sqrt{e^\alpha-1}) = \frac{\pi}{6}$.
Dividing by $2$,we get $\frac{\pi}{3} - \tan^{-1} \sqrt{e^\alpha-1} = \frac{\pi}{12}$,so $\tan^{-1} \sqrt{e^\alpha-1} = \frac{\pi}{3} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}$.
Thus,$\sqrt{e^\alpha-1} = \tan(\frac{\pi}{4}) = 1$,which means $e^\alpha-1 = 1$,so $e^\alpha = 2$.
Then $e^{-\alpha} = \frac{1}{2}$.
The quadratic equation with roots $2$ and $\frac{1}{2}$ is $(x-2)(x-\frac{1}{2}) = 0$,which simplifies to $x^2 - \frac{5}{2}x + 1 = 0$,or $2x^2 - 5x + 2 = 0$.

(C) Let $I = \int_0^{1/2} \frac{1+\sqrt{3}}{((x+1)^2(1-x)^6)^{1/4}} dx$.
Simplify the integrand: $((x+1)^2(1-x)^6)^{1/4} = (x+1)^{1/2}(1-x)^{3/2} = (1-x)^2 \left(\frac{x+1}{1-x}\right)^{1/2}$.
So,$I = \int_0^{1/2} \frac{1+\sqrt{3}}{\left(\frac{x+1}{1-x}\right)^{1/2} (1-x)^2} dx$.
Let $t = \frac{x+1}{1-x}$. Then $dt = \frac{(1-x)(1) - (x+1)(-1)}{(1-x)^2} dx = \frac{1-x+x+1}{(1-x)^2} dx = \frac{2}{(1-x)^2} dx$.
Thus,$\frac{dx}{(1-x)^2} = \frac{dt}{2}$.
When $x=0$,$t=1$. When $x=1/2$,$t = \frac{1.5}{0.5} = 3$.
$I = \int_1^3 \frac{1+\sqrt{3}}{\sqrt{t}} \cdot \frac{dt}{2} = \frac{1+\sqrt{3}}{2} \int_1^3 t^{-1/2} dt$.
$I = \frac{1+\sqrt{3}}{2} [2\sqrt{t}]_1^3 = (1+\sqrt{3})(\sqrt{3}-1) = 3-1 = 2$.

(A) Let $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \operatorname{cosec} x)^{17} dx$.
Substitute $\tan(\frac{x}{2}) = e^u$,then $\frac{1}{2} \sec^2(\frac{x}{2}) dx = e^u du$,so $dx = \frac{2 e^u}{1+e^{2u}} du$.
Also,$\operatorname{cosec} x = \frac{1}{\sin x} = \frac{1+e^{2u}}{2e^u} = \frac{e^u + e^{-u}}{2}$.
When $x = \frac{\pi}{4}$,$e^u = \tan(\frac{\pi}{8}) = \sqrt{2}-1$,so $u = \ln(\sqrt{2}-1) = -\ln(\sqrt{2}+1)$.
When $x = \frac{\pi}{2}$,$e^u = \tan(\frac{\pi}{4}) = 1$,so $u = 0$.
Substituting these into the integral:
$I = \int_{-\ln(\sqrt{2}+1)}^{0} (e^u + e^{-u})^{17} \cdot \frac{2}{e^u + e^{-u}} du = \int_{-\ln(\sqrt{2}+1)}^{0} 2(e^u + e^{-u})^{16} du$.
Since $f(u) = (e^u + e^{-u})^{16}$ is an even function,$\int_{-a}^{0} f(u) du = \int_{0}^{a} f(u) du$.
Thus,$I = \int_{0}^{\ln(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$.

(D) Given $\alpha = \int_0^1 e^{(9x + 3 \tan^{-1} x)} \left(\frac{12 + 9x^2}{1 + x^2}\right) dx$.
Let $t = 9x + 3 \tan^{-1} x$.
Then,$dt = \left(9 + \frac{3}{1 + x^2}\right) dx = \left(\frac{9(1 + x^2) + 3}{1 + x^2}\right) dx = \left(\frac{9 + 9x^2 + 3}{1 + x^2}\right) dx = \left(\frac{12 + 9x^2}{1 + x^2}\right) dx$.
When $x = 0$,$t = 9(0) + 3 \tan^{-1}(0) = 0$.
When $x = 1$,$t = 9(1) + 3 \tan^{-1}(1) = 9 + 3(\frac{\pi}{4}) = 9 + \frac{3\pi}{4}$.
Thus,$\alpha = \int_0^{9 + \frac{3\pi}{4}} e^t dt = [e^t]_0^{9 + \frac{3\pi}{4}} = e^{9 + \frac{3\pi}{4}} - e^0 = e^{9 + \frac{3\pi}{4}} - 1$.
Therefore,$1 + \alpha = e^{9 + \frac{3\pi}{4}}$.
Taking the natural logarithm,$\log_e |1 + \alpha| = 9 + \frac{3\pi}{4}$.
Finally,$\log_e |1 + \alpha| - \frac{3\pi}{4} = 9 + \frac{3\pi}{4} - \frac{3\pi}{4} = 9$.

(B) Let $I = \int_0^1 \frac{1}{2+\sqrt{x}} \, dx$.
Substitute $\sqrt{x} = t$,so $x = t^2$ and $dx = 2t \, dt$.
When $x = 0$,$t = 0$. When $x = 1$,$t = 1$.
Thus,$I = \int_0^1 \frac{2t}{2+t} \, dt$.
$I = 2 \int_0^1 \frac{t+2-2}{t+2} \, dt = 2 \int_0^1 \left(1 - \frac{2}{t+2}\right) \, dt$.
$I = 2 [t - 2 \log |t+2|]_0^1$.
$I = 2 [(1 - 2 \log 3) - (0 - 2 \log 2)]$.
$I = 2 [1 - 2 \log 3 + 2 \log 2] = 2 [1 + 2 \log(2/3)]$.
$I = 2 [\log e + \log(4/9)] = 2 \log(4e/9)$.
Therefore,the correct option is $B$.

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x(\tan ^5 x+\cot ^5 x)}$
Using $\sin 2x = 2 \sin x \cos x$ and $\cot x = \frac{1}{\tan x}$,we have:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{2 \sin x \cos x(\tan ^5 x+\frac{1}{\tan ^5 x})}$
Multiply numerator and denominator by $\sec^2 x$:
$I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec^2 x}{\tan x(\frac{\tan^{10} x+1}{\tan^5 x})} dx = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan^4 x \sec^2 x}{\tan^{10} x+1} dx$
Let $t = \tan^5 x$,then $dt = 5 \tan^4 x \sec^2 x dx$,so $\tan^4 x \sec^2 x dx = \frac{dt}{5}$.
When $x = \frac{\pi}{6}$,$t = (\frac{1}{\sqrt{3}})^5 = \frac{1}{9 \sqrt{3}}$. When $x = \frac{\pi}{4}$,$t = 1^5 = 1$.
$I = \frac{1}{2} \int_{\frac{1}{9 \sqrt{3}}}^{1} \frac{dt/5}{t^2+1} = \frac{1}{10} [\tan^{-1} t]_{\frac{1}{9 \sqrt{3}}}^{1}$
$I = \frac{1}{10} (\tan^{-1}(1) - \tan^{-1}(\frac{1}{9 \sqrt{3}})) = \frac{1}{10} (\frac{\pi}{4} - \tan^{-1}(\frac{1}{9 \sqrt{3}}))$

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$
$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$
$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\left(\frac{\sin x}{\cos x}\right)^{\frac{4}{3}} \cos^2 x} \, dx$
$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\tan^{\frac{4}{3}} x} \, dx$
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
When $x = \frac{\pi}{6}$,$t = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} = 3^{-\frac{1}{2}}$.
When $x = \frac{\pi}{3}$,$t = \tan(\frac{\pi}{3}) = \sqrt{3} = 3^{\frac{1}{2}}$.
$I = \int_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}} t^{-\frac{4}{3}} \, dt = \left[ \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} \right]_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}} = -3 \left[ t^{-\frac{1}{3}} \right]_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}}$
$= -3 \left( (3^{\frac{1}{2}})^{-\frac{1}{3}} - (3^{-\frac{1}{2}})^{-\frac{1}{3}} \right) = -3 \left( 3^{-\frac{1}{6}} - 3^{\frac{1}{6}} \right)$
$= 3 \cdot 3^{\frac{1}{6}} - 3 \cdot 3^{-\frac{1}{6}} = 3^{\frac{7}{6}} - 3^{\frac{5}{6}}$

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} dx$.
Substitute $t = \cos x$,then $dt = -\sin x dx$,or $\sin x dx = -dt$.
When $x = 0$,$t = \cos(0) = 1$.
When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$.
Substituting these into the integral:
$I = \int_1^0 \frac{-dt}{1+t^2} = \int_0^1 \frac{dt}{1+t^2}$.
Using the standard integral $\int \frac{1}{1+t^2} dt = \tan^{-1} t$:
$I = [\tan^{-1} t]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(A) Let $I = \int_{-3}^0 x \sqrt{x+4} \, dx$.
Substitute $x+4 = t$,so $x = t-4$ and $dx = dt$.
When $x = -3$,$t = 1$.
When $x = 0$,$t = 4$.
Substituting these into the integral:
$I = \int_1^4 (t-4) \sqrt{t} \, dt = \int_1^4 (t^{3/2} - 4t^{1/2}) \, dt$.
Integrating term by term:
$I = \left[ \frac{t^{5/2}}{5/2} - 4 \cdot \frac{t^{3/2}}{3/2} \right]_1^4 = \left[ \frac{2}{5} t^{5/2} - \frac{8}{3} t^{3/2} \right]_1^4$.
Evaluating at the limits:
$I = \left( \frac{2}{5}(4)^{5/2} - \frac{8}{3}(4)^{3/2} \right) - \left( \frac{2}{5}(1)^{5/2} - \frac{8}{3}(1)^{3/2} \right)$.
$I = \left( \frac{2}{5}(32) - \frac{8}{3}(8) \right) - \left( \frac{2}{5} - \frac{8}{3} \right)$.
$I = \left( \frac{64}{5} - \frac{64}{3} \right) - \left( \frac{6-40}{15} \right) = \left( \frac{192-320}{15} \right) - \left( \frac{-34}{15} \right) = \frac{-128}{15} + \frac{34}{15} = -\frac{94}{15}$.

(B) To evaluate the integral $I = \int_{2}^{3} \frac{x}{x^{2}-1} d x$,let $u = x^{2}-1$.
Then $du = 2x \, dx$,which implies $x \, dx = \frac{1}{2} du$.
When $x = 2$,$u = 2^{2}-1 = 3$.
When $x = 3$,$u = 3^{2}-1 = 8$.
Substituting these into the integral:
$I = \int_{3}^{8} \frac{1}{u} \cdot \frac{1}{2} du$
$I = \frac{1}{2} [\log |u|]_{3}^{8}$
$I = \frac{1}{2} (\log 8 - \log 3)$
$I = \frac{1}{2} \log \left(\frac{8}{3}\right)$.

(A) Let $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.
Substitute $u = 1 + \log x$. Then $du = \frac{1}{x} dx$.
When $x = 1$,$u = 1 + \log 1 = 1 + 0 = 1$.
When $x = e$,$u = 1 + \log e = 1 + 1 = 2$.
Thus,$I = \int_{1}^{2} \frac{du}{u^{2}} = \int_{1}^{2} u^{-2} du$.
$I = \left[ \frac{u^{-1}}{-1} \right]_{1}^{2} = \left[ -\frac{1}{u} \right]_{1}^{2}$.
$I = -\left( \frac{1}{2} - \frac{1}{1} \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2}$.

(C) Let $I = \int_0^4 \frac{1}{1+\sqrt{x}} \, dx$.
Substitute $x = t^2$,so $dx = 2t \, dt$.
When $x = 0, t = 0$ and when $x = 4, t = 2$.
Then,$I = \int_0^2 \frac{2t}{1+t} \, dt$.
$I = 2 \int_0^2 \frac{(1+t)-1}{1+t} \, dt$.
$I = 2 \int_0^2 \left(1 - \frac{1}{1+t}\right) \, dt$.
$I = 2 [t - \ln(1+t)]_0^2$.
$I = 2 [(2 - \ln 3) - (0 - \ln 1)]$.
$I = 2(2 - \ln 3) = 4 - 2\ln 3 = 4 - \ln(3^2) = 4 - \ln 9$.
Since $4 = \ln(e^4)$,we have $I = \ln(e^4) - \ln 9 = \ln \left(\frac{e^4}{9}\right)$.

(C) Let $I = \int_0^{\frac{\pi}{4}} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} \, dx$.
Divide the numerator and denominator by $\cos^6 x$:
$I = \int_0^{\frac{\pi}{4}} \frac{\tan^2 x \sec^2 x}{(1 + \tan^3 x)^2} \, dx$.
Let $u = \tan^3 x$,then $du = 3 \tan^2 x \sec^2 x \, dx$,which implies $\tan^2 x \sec^2 x \, dx = \frac{du}{3}$.
When $x = 0$,$u = 0$. When $x = \frac{\pi}{4}$,$u = 1$.
Substituting these into the integral:
$I = \int_0^1 \frac{1}{3(1 + u)^2} \, du = \frac{1}{3} \left[ -\frac{1}{1 + u} \right]_0^1$.
$I = \frac{1}{3} \left( -\frac{1}{2} - (-1) \right) = \frac{1}{3} \left( \frac{1}{2} \right) = \frac{1}{6}$.

(C) Let $1+\tan x = t$. Then $\sec^2 x \, dx = dt$.
When $x = 0$,$t = 1+\tan(0) = 1$.
When $x = \frac{\pi}{4}$,$t = 1+\tan(\frac{\pi}{4}) = 2$.
The integral becomes:
$\int_1^2 \frac{dt}{t(t+1)}$.
Using partial fractions: $\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
So,the integral is $\int_1^2 \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$.
$= [\log|t| - \log|t+1|]_1^2 = [\log|\frac{t}{t+1}|]_1^2$.
$= \log(\frac{2}{3}) - \log(\frac{1}{2}) = \log(\frac{2/3}{1/2}) = \log(\frac{4}{3})$.

(B) Let $I = \int_0^3 \frac{dx}{(x+2) \sqrt{x+1}}$.
Substitute $t = \sqrt{x+1}$,so $t^2 = x+1$ and $x = t^2 - 1$.
Then $dx = 2t \, dt$.
When $x = 0$,$t = 1$. When $x = 3$,$t = 2$.
The integral becomes $I = \int_1^2 \frac{2t \, dt}{(t^2 - 1 + 2) t} = \int_1^2 \frac{2 \, dt}{t^2 + 1}$.
Integrating,we get $I = [2 \tan^{-1}(t)]_1^2 = 2(\tan^{-1}(2) - \tan^{-1}(1))$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,
$I = 2 \tan^{-1}\left(\frac{2-1}{1+2(1)}\right) = 2 \tan^{-1}\left(\frac{1}{3}\right)$.

(D) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x$.
Substitute $t = \operatorname{cosec} x$,then $dt = -\operatorname{cosec} x \cot x \, dx$,which implies $\operatorname{cosec} x \cot x \, dx = -dt$.
When $x = \frac{\pi}{6}$,$t = \operatorname{cosec}(\frac{\pi}{6}) = 2$.
When $x = \frac{\pi}{2}$,$t = \operatorname{cosec}(\frac{\pi}{2}) = 1$.
Thus,$I = \int_{2}^{1} \frac{-dt}{1+t^2} = \int_{1}^{2} \frac{dt}{1+t^2}$.
Evaluating the integral,we get $I = [\tan^{-1} t]_{1}^{2} = \tan^{-1}(2) - \tan^{-1}(1)$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1}(\frac{x-y}{1+xy})$,we have $I = \tan^{-1}(\frac{2-1}{1+2 \cdot 1}) = \tan^{-1}(\frac{1}{3})$.

(C) Let $I = \int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}}$.
Substitute $u = \sqrt{3x+2}$. Then $u^2 = 3x+2$,so $2u du = 3 dx$,which means $dx = \frac{2}{3} u du$.
When $x=0$,$u = \sqrt{2}$. When $x=1$,$u = \sqrt{5}$.
Substituting these into the integral:
$I = \int_{\sqrt{2}}^{\sqrt{5}} \frac{\frac{2}{3} u du}{u^2+u} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{u du}{u(u+1)} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{du}{u+1}$.
Integrating,we get $I = \frac{2}{3} [\log |u+1|]_{\sqrt{2}}^{\sqrt{5}}$.
$I = \frac{2}{3} (\log |\sqrt{5}+1| - \log |\sqrt{2}+1|) = \frac{2}{3} \log \left|\frac{\sqrt{5}+1}{\sqrt{2}+1}\right|$.
Thus,the correct option is $C$.

(D) Let $I = \int_{\frac{1}{3}}^1 \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx$.
Factor out $x^3$ from the term inside the cube root:
$I = \int_{\frac{1}{3}}^1 \frac{(x^3(\frac{1}{x^2}-1))^{\frac{1}{3}}}{x^4} dx = \int_{\frac{1}{3}}^1 \frac{x(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^4} dx = \int_{\frac{1}{3}}^1 \frac{(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^3} dx$.
Let $u = \frac{1}{x^2} - 1$. Then $du = -\frac{2}{x^3} dx$,which implies $\frac{dx}{x^3} = -\frac{1}{2} du$.
When $x = \frac{1}{3}$,$u = \frac{1}{(1/3)^2} - 1 = 9 - 1 = 8$.
When $x = 1$,$u = \frac{1}{1^2} - 1 = 0$.
Substituting these into the integral:
$I = \int_8^0 u^{\frac{1}{3}} (-\frac{1}{2}) du = \frac{1}{2} \int_0^8 u^{\frac{1}{3}} du$.
$I = \frac{1}{2} [\frac{u^{\frac{4}{3}}}{4/3}]_0^8 = \frac{1}{2} \cdot \frac{3}{4} [u^{\frac{4}{3}}]_0^8 = \frac{3}{8} (8^{\frac{4}{3}} - 0) = \frac{3}{8} (16) = 6$.
Thus,the correct option is $D$.

(B) Let $I = \frac{1}{2} \int_2^3 \frac{2x}{x^2+1} dx$.
Using the substitution method,let $u = x^2+1$.
Then $du = 2x dx$.
When $x=2$,$u = 2^2+1 = 5$.
When $x=3$,$u = 3^2+1 = 10$.
Substituting these into the integral:
$I = \frac{1}{2} \int_5^{10} \frac{1}{u} du$
$I = \frac{1}{2} [\log |u|]_5^{10}$
$I = \frac{1}{2} (\log 10 - \log 5)$
$I = \frac{1}{2} \log \left(\frac{10}{5}\right)$
$I = \frac{1}{2} \log (2)$.

(A) Let $I = k \int_{0}^{1} x \cdot f(3x) \, dx$.
Substitute $t = 3x$,then $dt = 3 \, dx$,which implies $dx = \frac{dt}{3}$.
When $x = 0$,$t = 0$. When $x = 1$,$t = 3$.
Substituting these into the integral:
$I = k \int_{0}^{3} \left(\frac{t}{3}\right) \cdot f(t) \cdot \left(\frac{dt}{3}\right)$
$I = \frac{k}{9} \int_{0}^{3} t \cdot f(t) \, dt$.
Given that $I = \int_{0}^{3} t \cdot f(t) \, dt$,we equate the two expressions:
$\frac{k}{9} \int_{0}^{3} t \cdot f(t) \, dt = \int_{0}^{3} t \cdot f(t) \, dt$.
Comparing the coefficients,we get $\frac{k}{9} = 1$,which implies $k = 9$.

(D) Given,$\int_{0}^{1} f(x) dx = 5$.
Let $I = 100 \int_{0}^{1} x^{9} f(x^{10}) dx$.
Substitute $x^{10} = t$,then $10x^{9} dx = dt$,which implies $x^{9} dx = \frac{dt}{10}$.
When $x = 0, t = 0$ and when $x = 1, t = 1$.
Therefore,$I = 100 \int_{0}^{1} f(t) \frac{dt}{10} = 10 \int_{0}^{1} f(t) dt$.
Since $\int_{0}^{1} f(t) dt = \int_{0}^{1} f(x) dx = 5$,we have $I = 10 \times 5 = 50$.
Thus,the required value is $\int_{0}^{1} f(x) dx + I = 5 + 50 = 55$.

(B) Let $I = \int_{1/5}^{1/2} \frac{\sqrt{x-x^2}}{x^3} dx$.
We can rewrite the integrand as $\frac{\sqrt{x^2(\frac{1}{x}-1)}}{x^3} = \frac{x\sqrt{\frac{1}{x}-1}}{x^3} = \frac{\sqrt{\frac{1}{x}-1}}{x^2}$.
Let $u = \frac{1}{x} - 1$. Then $du = -\frac{1}{x^2} dx$,which implies $-\frac{1}{x^2} dx = du$.
When $x = 1/5$,$u = 5 - 1 = 4$.
When $x = 1/2$,$u = 2 - 1 = 1$.
Substituting these into the integral: $I = \int_{4}^{1} \sqrt{u} (-du) = \int_{1}^{4} u^{1/2} du$.
Evaluating the integral: $I = [\frac{u^{3/2}}{3/2}]_{1}^{4} = \frac{2}{3} [u^{3/2}]_{1}^{4}$.
$I = \frac{2}{3} (4^{3/2} - 1^{3/2}) = \frac{2}{3} (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3}$.

(D) Let $I = \int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x$.
Substitute $t = \frac{3}{x}$.
Then $dt = -\frac{3}{x^2} dx$,which implies $\frac{1}{x^2} dx = -\frac{1}{3} dt$.
When $x = 3$,$t = \frac{3}{3} = 1$.
When $x = \frac{1}{25}$,$t = \frac{3}{1/25} = 75$.
Substituting these into the integral:
$I = \int_{75}^1 e^t \left(-\frac{1}{3} dt\right) = -\frac{1}{3} \int_{75}^1 e^t dt$.
$I = -\frac{1}{3} [e^t]_{75}^1 = -\frac{1}{3} (e^1 - e^{75})$.
$I = \frac{1}{3} (e^{75} - e)$.

(D) Given $f(x) = x^2$ and $g(x) = \cos x$,we have $g(f(x)) = \cos(x^2)$.
For the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$,the roots are given by the quadratic formula:
$x = \frac{9\pi \pm \sqrt{(9\pi)^2 - 4(18)(\pi^2)}}{2(18)} = \frac{9\pi \pm \sqrt{81\pi^2 - 72\pi^2}}{36} = \frac{9\pi \pm 3\pi}{36}$.
Thus,the roots are $x = \frac{12\pi}{36} = \frac{\pi}{3}$ and $x = \frac{6\pi}{36} = \frac{\pi}{6}$.
Since $\alpha < \beta$,we have $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$.
We need to evaluate $I = \int_{\pi/6}^{\pi/3} x \cos(x^2) dx$.
Let $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.
When $x = \frac{\pi}{6}$,$t = \frac{\pi^2}{36}$. When $x = \frac{\pi}{3}$,$t = \frac{\pi^2}{9}$.
Substituting these into the integral:
$I = \int_{\pi^2/36}^{\pi^2/9} \cos(t) \frac{dt}{2} = \frac{1}{2} [\sin(t)]_{\pi^2/36}^{\pi^2/9} = \frac{1}{2} (\sin \frac{\pi^2}{9} - \sin \frac{\pi^2}{36})$.

(D) Let $I = \int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{1/3}} dx$.
Substitute $t = \log x$,then $dt = \frac{1}{x} dx$.
When $x = \frac{1}{e}$,$t = \log(\frac{1}{e}) = -1$.
When $x = \frac{2}{e}$,$t = \log(\frac{2}{e}) = \log 2 - \log e = \log 2 - 1$.
Thus,$I = \int_{\log 2 - 1}^{-1} t^{-1/3} dt$.
Integrating,we get $I = \left[ \frac{t^{2/3}}{2/3} \right]_{\log 2 - 1}^{-1} = \frac{3}{2} \left[ t^{2/3} \right]_{\log 2 - 1}^{-1}$.
$I = \frac{3}{2} \left[ (-1)^{2/3} - (\log 2 - 1)^{2/3} \right]$.
Since $(-1)^{2/3} = ((-1)^2)^{1/3} = 1^{1/3} = 1$,we have $I = \frac{3}{2} \left[ 1 - (\log 2 - 1)^{2/3} \right]$.

(A) Let $I = \int_1^{e^2} \frac{dx}{x(1+\log x)^2}$.
Substitute $t = 1 + \log x$. Then $dt = \frac{1}{x} dx$.
When $x = 1$,$t = 1 + \log(1) = 1 + 0 = 1$.
When $x = e^2$,$t = 1 + \log(e^2) = 1 + 2 = 3$.
Thus,$I = \int_1^3 \frac{dt}{t^2} = \int_1^3 t^{-2} dt$.
Integrating,we get $I = \left[ \frac{t^{-1}}{-1} \right]_1^3 = \left[ -\frac{1}{t} \right]_1^3$.
Evaluating the limits,$I = \left( -\frac{1}{3} \right) - \left( -\frac{1}{1} \right) = -\frac{1}{3} + 1 = \frac{2}{3}$.

(C) Let $I = \int_0^{\pi / 2} e^{\sin x} \cos x \, dx$.
Substitute $\sin x = t$. Then,differentiating both sides with respect to $x$,we get $\cos x \, dx = dt$.
Now,change the limits of integration:
When $x = 0$,$t = \sin(0) = 0$.
When $x = \frac{\pi}{2}$,$t = \sin(\frac{\pi}{2}) = 1$.
Substituting these into the integral,we get:
$I = \int_0^1 e^t \, dt$.
The integral of $e^t$ is $e^t$.
$I = [e^t]_0^1 = e^1 - e^0 = e - 1$.

(A) Given integral: $\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6}$.
Let $e^t - 1 = u^2$,then $e^t dt = 2u du$,so $dt = \frac{2u du}{u^2+1}$.
When $t = \log_e 2$,$u = \sqrt{e^{\log_e 2} - 1} = \sqrt{2-1} = 1$.
When $t = x$,$u = \sqrt{e^x - 1}$.
The integral becomes $\int_{1}^{\sqrt{e^x-1}} \frac{2u du}{(u^2+1)u} = 2 \int_{1}^{\sqrt{e^x-1}} \frac{du}{u^2+1} = 2 [\tan^{-1} u]_{1}^{\sqrt{e^x-1}} = \frac{\pi}{6}$.
$2 [\tan^{-1}(\sqrt{e^x-1}) - \tan^{-1}(1)] = \frac{\pi}{6}$.
$\tan^{-1}(\sqrt{e^x-1}) - \frac{\pi}{4} = \frac{\pi}{12}$.
$\tan^{-1}(\sqrt{e^x-1}) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi+3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.
$\sqrt{e^x-1} = \tan(\frac{\pi}{3}) = \sqrt{3}$.
$e^x - 1 = 3 \Rightarrow e^x = 4$.
$x = \log_e 4 = \log_e(2^2) = 2 \log_e 2$.
Thus,option $A$ is correct.

(A) Let $I = \int_{8}^{18} \frac{1}{(x+2) \sqrt{x-3}} \, dx$.
Substitute $t = \sqrt{x-3}$,so $t^2 = x-3$,which implies $x = t^2+3$ and $dx = 2t \, dt$.
When $x = 8$,$t = \sqrt{8-3} = \sqrt{5}$.
When $x = 18$,$t = \sqrt{18-3} = \sqrt{15}$.
Substituting these into the integral:
$I = \int_{\sqrt{5}}^{\sqrt{15}} \frac{2t \, dt}{(t^2+3+2)t} = \int_{\sqrt{5}}^{\sqrt{15}} \frac{2 \, dt}{t^2+5}$.
Using the formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = 2 \left[ \frac{1}{\sqrt{5}} \tan^{-1}(\frac{t}{\sqrt{5}}) \right]_{\sqrt{5}}^{\sqrt{15}}$.
$I = \frac{2}{\sqrt{5}} \left[ \tan^{-1}(\frac{\sqrt{15}}{\sqrt{5}}) - \tan^{-1}(\frac{\sqrt{5}}{\sqrt{5}}) \right]$.
$I = \frac{2}{\sqrt{5}} \left[ \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \right]$.
$I = \frac{2}{\sqrt{5}} \left[ \frac{\pi}{3} - \frac{\pi}{4} \right] = \frac{2}{\sqrt{5}} \left[ \frac{\pi}{12} \right] = \frac{\pi}{6 \sqrt{5}}$.

(B) Let $I = \int_0^{\pi / 4} \frac{\sec x}{3 \cos x+4 \sin x} d x$.
Multiply the numerator and denominator by $\sec x$:
$I = \int_0^{\pi / 4} \frac{\sec^2 x}{3 + 4 \tan x} d x$.
Let $u = 3 + 4 \tan x$. Then $du = 4 \sec^2 x d x$,so $\sec^2 x d x = \frac{du}{4}$.
When $x = 0$,$u = 3 + 4(0) = 3$.
When $x = \pi / 4$,$u = 3 + 4(1) = 7$.
Substituting these into the integral:
$I = \int_3^7 \frac{1}{u} \cdot \frac{du}{4} = \frac{1}{4} [\log |u|]_3^7$.
$I = \frac{1}{4} (\log 7 - \log 3) = \frac{1}{4} \log \left(\frac{7}{3}\right)$.

(A) Let $I = \int_1^2 x \sqrt{4-x^2} \, dx$.
Substitute $u = 4-x^2$,then $du = -2x \, dx$,which implies $x \, dx = -\frac{1}{2} \, du$.
When $x = 1$,$u = 4 - (1)^2 = 3$.
When $x = 2$,$u = 4 - (2)^2 = 0$.
Substituting these into the integral:
$I = \int_3^0 \sqrt{u} \left(-\frac{1}{2}\right) \, du = \frac{1}{2} \int_0^3 u^{1/2} \, du$.
$I = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_0^3 = \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_0^3$.
$I = \frac{1}{3} (3^{3/2} - 0) = \frac{1}{3} (3 \sqrt{3}) = \sqrt{3}$.

(B) Let $I = \int_0^{\frac{\pi}{2}} \sin^4 \theta \cos^3 \theta \, d\theta$.
We can write $\cos^3 \theta \, d\theta$ as $\cos^2 \theta \cdot \cos \theta \, d\theta = (1 - \sin^2 \theta) \cos \theta \, d\theta$.
Substituting this into the integral,we get:
$I = \int_0^{\frac{\pi}{2}} \sin^4 \theta (1 - \sin^2 \theta) \cos \theta \, d\theta$.
Let $t = \sin \theta$,then $dt = \cos \theta \, d\theta$.
When $\theta = 0$,$t = 0$. When $\theta = \frac{\pi}{2}$,$t = 1$.
$I = \int_0^1 t^4 (1 - t^2) \, dt = \int_0^1 (t^4 - t^6) \, dt$.
Integrating term by term:
$I = \left[ \frac{t^5}{5} - \frac{t^7}{7} \right]_0^1 = \left( \frac{1}{5} - \frac{1}{7} \right) - (0 - 0) = \frac{7 - 5}{35} = \frac{2}{35}$.

(C) Given that $\frac{d}{dt}(t \log t - t) = \log t$.
Let $I = \int_0^1 2x \log(1+x^2) dx$.
Substitute $t = 1+x^2$,then $dt = 2x dx$.
When $x=0$,$t=1$. When $x=1$,$t=2$.
Thus,$I = \int_1^2 \log t dt$.
Using the given derivative,$\int \log t dt = t \log t - t + C$.
So,$I = [t \log t - t]_1^2 = (2 \log 2 - 2) - (1 \log 1 - 1) = 2 \log 2 - 2 - 0 + 1 = \log 4 - 1$.
Therefore,$\exp(I) = \exp(\log 4 - 1) = e^{\log 4} \cdot e^{-1} = 4 \cdot \frac{1}{e} = \frac{4}{e}$.
Thus,the correct option is $C$.

(C, D) For $I_{1}=\int_{-2}^{2} \frac{dx}{4+x^{2}}$,the integrand is positive,so $I_{1} > 0$.
If we substitute $x=\frac{1}{t}$,then $dx = -\frac{1}{t^{2}} dt$. The limits change from $x=-2$ to $t=-1/2$ and $x=2$ to $t=1/2$.
$I_{1} = \int_{-1/2}^{1/2} \frac{-dt/t^{2}}{4+1/t^{2}} = \int_{-1/2}^{1/2} \frac{-dt}{4t^{2}+1}$. Since the integrand is positive,the integral must be positive,but this substitution leads to a negative value,which is incorrect due to the discontinuity of $1/t$ at $t=0$. Thus,it is not possible.
For $I_{2}=\int_{0}^{1} \sqrt{x^{2}+1} dx$,if we put $x=\operatorname{cosec} \theta$,then $\operatorname{cosec} \theta \in (-\infty, -1] \cup [1, \infty)$. Since $x \in (0, 1)$,the range of $x$ does not match the range of $\operatorname{cosec} \theta$. Thus,it is not possible.
Therefore,options $C$ and $D$ are correct.

(B) Given that $2^{1-a} + 2^{1+a}$,$f(a)$,and $3^a + 3^{-a}$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.
Using the $AM$-$GM$ inequality,$2^{1-a} + 2^{1+a} = 2(2^{-a} + 2^a) \geq 2(2) = 4$ and $3^a + 3^{-a} \geq 2$. Both reach their minimum at $a=0$.
Thus,the minimum value $\alpha$ of $f(a)$ is $\alpha = \frac{1}{2}(4 + 2) = 3$.
The integral becomes $I = \int_{\log_e(2)}^{\log_e(3)} \frac{dx}{e^{2x} - e^{-2x}} = \int_{\log_e(2)}^{\log_e(3)} \frac{e^{2x} dx}{e^{4x} - 1}$.
Let $t = e^{2x}$,then $dt = 2e^{2x} dx$,so $e^{2x} dx = \frac{dt}{2}$.
When $x = \log_e(2)$,$t = e^{2\log_e(2)} = 4$. When $x = \log_e(3)$,$t = e^{2\log_e(3)} = 9$.
$I = \frac{1}{2} \int_{4}^{9} \frac{dt}{t^2 - 1} = \frac{1}{2} \cdot \frac{1}{2} [\log_e|\frac{t-1}{t+1}|]_{4}^{9} = \frac{1}{4} [\log_e(\frac{8}{10}) - \log_e(\frac{3}{5})] = \frac{1}{4} \log_e(\frac{8/10}{3/5}) = \frac{1}{4} \log_e(\frac{4}{3})$.