Fundamental definite integration Questions in English

(C) Let $y = \mathop {\lim }\limits_{x \to \pi /2} \frac{{\int_{\pi /2}^x {t\,dt} }}{{\sin (2x - \pi )}}$.
Using the Fundamental Theorem of Calculus,$\int_{\pi /2}^x t \, dt = \left[ \frac{t^2}{2} \right]_{\pi /2}^x = \frac{x^2}{2} - \frac{\pi^2}{8}$.
So,$y = \mathop {\lim }\limits_{x \to \pi /2} \frac{\frac{x^2}{2} - \frac{\pi^2}{8}}{\sin(2x - \pi )} = \mathop {\lim }\limits_{x \to \pi /2} \frac{4x^2 - \pi^2}{8 \sin(2x - \pi )}$.
Factoring the numerator: $4x^2 - \pi^2 = (2x - \pi)(2x + \pi)$.
$y = \frac{1}{8} \mathop {\lim }\limits_{x \to \pi /2} \frac{(2x - \pi)(2x + \pi)}{\sin(2x - \pi )}$.
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we have $\mathop {\lim }\limits_{x \to \pi /2} \frac{\sin(2x - \pi)}{2x - \pi} = 1$.
Therefore,$y = \frac{1}{8} \times (2(\frac{\pi}{2}) + \pi) \times 1 = \frac{1}{8} \times (2\pi) = \frac{\pi}{4}$.

(D) Let $f(x) = \int_0^x {t{e^{ - {t^2}}}} dt$.
By the Fundamental Theorem of Calculus,$f'(x) = x{e^{ - {x^2}}}$.
To find the critical points,set $f'(x) = 0$,which gives $x{e^{ - {x^2}}} = 0$. Since ${e^{ - {x^2}}} \neq 0$ for any real $x$,we have $x = 0$.
Now,find the second derivative: $f''(x) = \frac{d}{dx}(x{e^{ - {x^2}}}) = {e^{ - {x^2}}} + x({e^{ - {x^2}}}(-2x)) = {e^{ - {x^2}}}(1 - 2{x^2})$.
Evaluating at $x = 0$: $f''(0) = {e^0}(1 - 0) = 1$.
Since $f''(0) > 0$,the function has a local minimum at $x = 0$.
The minimum value is $f(0) = \int_0^0 {t{e^{ - {t^2}}}} dt = 0$.

(C) We are given the integral $I = \int_0^1 {{e^{2\ln x}}dx}$.
Using the logarithmic property $a \ln b = \ln(b^a)$,we can rewrite the exponent:
$2\ln x = \ln(x^2)$.
Substituting this into the integral,we get:
$I = \int_0^1 {{e^{\ln(x^2)}}dx}$.
Since $e^{\ln(f(x))} = f(x)$,the integral simplifies to:
$I = \int_0^1 {{x^2}dx}$.
Now,we evaluate the definite integral:
$I = \left[ \frac{x^3}{3} \right]_0^1$.
Applying the limits:
$I = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}$.

(A) We know that $\tan^2 x = \sec^2 x - 1$.
Substituting this into the integral:
$\int_0^{\pi /4} \tan^2 x \, dx = \int_0^{\pi /4} (\sec^2 x - 1) \, dx$
$= \int_0^{\pi /4} \sec^2 x \, dx - \int_0^{\pi /4} 1 \, dx$
$= [\tan x]_0^{\pi /4} - [x]_0^{\pi /4}$
$= (\tan(\frac{\pi}{4}) - \tan(0)) - (\frac{\pi}{4} - 0)$
$= (1 - 0) - \frac{\pi}{4}$
$= 1 - \frac{\pi}{4}$.

(C) Let $I = \int_0^{\pi /2} \frac{x + \sin x}{1 + \cos x} \, dx$.
Using the identity $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get:
$I = \int_0^{\pi /2} \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \, dx$
$I = \int_0^{\pi /2} \left( \frac{x}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) \, dx$
Consider the integral of the first term using integration by parts:
$\int \frac{x}{2} \sec^2 \frac{x}{2} \, dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} \, dx$
Substituting this back into the expression for $I$:
$I = \left[ x \tan \frac{x}{2} - \int \tan \frac{x}{2} \, dx + \int \tan \frac{x}{2} \, dx \right]_0^{\pi /2}$
$I = \left[ x \tan \frac{x}{2} \right]_0^{\pi /2}$
$I = \frac{\pi}{2} \tan \frac{\pi}{4} - 0 \cdot \tan 0 = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}$.

(B) Let $I = \int_0^{\pi /2} e^x \sin x \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = \sin x$ and $dv = e^x \, dx$. Then $du = \cos x \, dx$ and $v = e^x$.
$I = [e^x \sin x]_0^{\pi /2} - \int_0^{\pi /2} e^x \cos x \, dx$.
Applying integration by parts again to $\int e^x \cos x \, dx$:
Let $u = \cos x$ and $dv = e^x \, dx$. Then $du = -\sin x \, dx$ and $v = e^x$.
$I = [e^x \sin x]_0^{\pi /2} - ([e^x \cos x]_0^{\pi /2} - \int_0^{\pi /2} e^x (-\sin x) \, dx)$.
$I = [e^x \sin x]_0^{\pi /2} - [e^x \cos x]_0^{\pi /2} - I$.
$2I = [e^x(\sin x - \cos x)]_0^{\pi /2}$.
$2I = (e^{\pi /2}(\sin(\pi /2) - \cos(\pi /2))) - (e^0(\sin 0 - \cos 0))$.
$2I = (e^{\pi /2}(1 - 0)) - (1(0 - 1))$.
$2I = e^{\pi /2} + 1$.
$I = \frac{1}{2}(e^{\pi /2} + 1)$.

(A) Let $I = \int_0^{\pi /2} \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \,dx$.
Substitute $\sin x = t$,then $\cos x \,dx = dt$.
When $x = 0$,$t = 0$. When $x = \pi/2$,$t = 1$.
The integral becomes $I = \int_0^1 \frac{1}{(1 + t)(2 + t)} \,dt$.
Using partial fractions,$\frac{1}{(1 + t)(2 + t)} = \frac{1}{1 + t} - \frac{1}{2 + t}$.
Thus,$I = \int_0^1 \left( \frac{1}{1 + t} - \frac{1}{2 + t} \right) \,dt$.
$I = [\log |1 + t| - \log |2 + t|]_0^1$.
$I = [\log \frac{1 + t}{2 + t}]_0^1$.
$I = \log \frac{2}{3} - \log \frac{1}{2} = \log \left( \frac{2/3}{1/2} \right) = \log \frac{4}{3}$.

(D) Let $I = \int_1^2 \frac{1}{x^2} e^{-\frac{1}{x}} \, dx$.
Substitute $t = -\frac{1}{x}$.
Then,$dt = \frac{1}{x^2} \, dx$.
When $x = 1$,$t = -1$.
When $x = 2$,$t = -\frac{1}{2}$.
Substituting these into the integral:
$I = \int_{-1}^{-1/2} e^t \, dt = [e^t]_{-1}^{-1/2}$.
$I = e^{-1/2} - e^{-1} = \frac{1}{\sqrt{e}} - \frac{1}{e}$.
$I = \frac{\sqrt{e} - 1}{e}$.

(A) Let $I = \int_0^1 {{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx$.
Substitute $x = \tan \theta$,so $dx = \sec^2 \theta \, d\theta$.
When $x = 0$,$\theta = 0$. When $x = 1$,$\theta = \frac{\pi}{4}$.
Since $\sin^{-1}(\frac{2\tan \theta}{1+\tan^2 \theta}) = \sin^{-1}(\sin 2\theta) = 2\theta$ for $x \in [0, 1]$,the integral becomes:
$I = \int_0^{\pi/4} 2\theta \sec^2 \theta \, d\theta$.
Using integration by parts:
$I = 2 [\theta \tan \theta]_0^{\pi/4} - 2 \int_0^{\pi/4} \tan \theta \, d\theta$.
$I = 2 [\frac{\pi}{4} \cdot 1 - 0] - 2 [\ln |\sec \theta|]_0^{\pi/4}$.
$I = \frac{\pi}{2} - 2 \ln(\sec \frac{\pi}{4}) = \frac{\pi}{2} - 2 \ln(\sqrt{2})$.
Thus,the correct option is $A$.

(D) We know that $\int \text{cosec} \, ax \, dx = \frac{1}{a} \log |\tan(\frac{ax}{2})| + C$.
Applying this to the given integral:
$\int_{\pi /6}^{\pi /4} \text{cosec} \, 2x \, dx = \left[ \frac{1}{2} \log |\tan(\frac{2x}{2})| \right]_{\pi /6}^{\pi /4}$
$= \frac{1}{2} [\log \tan x]_{\pi /6}^{\pi /4}$
$= \frac{1}{2} [\log \tan(\frac{\pi}{4}) - \log \tan(\frac{\pi}{6})]$
$= \frac{1}{2} [\log(1) - \log(\frac{1}{\sqrt{3}})]$
$= \frac{1}{2} [0 - \log(3^{-1/2})]$
$= \frac{1}{2} [\frac{1}{2} \log 3] = \frac{1}{4} \log 3 = \frac{1}{2} \log \sqrt{3}$.

(B) Let $I = \int_0^{\pi /2} \sqrt{\cos \theta} \sin^3 \theta \, d\theta$.
Substitute $t = \cos \theta$,then $dt = -\sin \theta \, d\theta$.
When $\theta = 0$,$t = 1$. When $\theta = \pi/2$,$t = 0$.
Thus,$I = \int_1^0 \sqrt{t} (1 - t^2) (-dt) = \int_0^1 (t^{1/2} - t^{5/2}) \, dt$.
Integrating term by term,we get $I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} \right]_0^1 = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} \right]_0^1$.
Evaluating at the limits,$I = \frac{2}{3} - \frac{2}{7} = \frac{14 - 6}{21} = \frac{8}{21}$.

(C) Let $I = \int_a^b \frac{\log x}{x} \, dx$.
Substitute $u = \log x$,then $du = \frac{1}{x} \, dx$.
When $x = a$,$u = \log a$.
When $x = b$,$u = \log b$.
Therefore,$I = \int_{\log a}^{\log b} u \, du$.
$I = \left[ \frac{u^2}{2} \right]_{\log a}^{\log b} = \frac{1}{2} [(\log b)^2 - (\log a)^2]$.
Using the identity $x^2 - y^2 = (x + y)(x - y)$,we get:
$I = \frac{1}{2} [(\log b + \log a)(\log b - \log a)]$.
Since $\log b + \log a = \log (ab)$ and $\log b - \log a = \log \left( \frac{b}{a} \right)$,
$I = \frac{1}{2} \log (ab) \log \left( \frac{b}{a} \right)$.

(A) Let $I = \int_0^1 {{\tan ^{ - 1}}x\,dx}$.
Using integration by parts,$\int {u,dv = uv - \int {v,du} } $.
Let $u = {\tan ^{ - 1}}x$ and $dv = dx$.
Then $du = \frac{1}{{1 + {x^2}}}dx$ and $v = x$.
$I = [x{\tan ^{ - 1}}x]_0^1 - \int_0^1 {\frac{x}{{1 + {x^2}}}dx} $.
$I = (1 \cdot {\tan ^{ - 1}}(1) - 0 \cdot {\tan ^{ - 1}}(0)) - \frac{1}{2}\int_0^1 {\frac{{2x}}{{1 + {x^2}}}dx} $.
$I = \frac{\pi }{4} - \frac{1}{2}[\log (1 + {x^2})]_0^1$.
$I = \frac{\pi }{4} - \frac{1}{2}(\log 2 - \log 1)$.
Since $\log 1 = 0$,we get $I = \frac{\pi }{4} - \frac{1}{2}\log 2$.

(D) Let $I = \int_0^1 \frac{dx}{[(a - b)x + b]^2}$.
Substitute $t = (a - b)x + b$,then $dt = (a - b)dx$,which implies $dx = \frac{dt}{a - b}$.
When $x = 0$,$t = b$. When $x = 1$,$t = a$.
Substituting these into the integral:
$I = \int_b^a \frac{1}{t^2} \cdot \frac{dt}{a - b} = \frac{1}{a - b} \int_b^a t^{-2} dt$.
Evaluating the integral:
$I = \frac{1}{a - b} \left[ -\frac{1}{t} \right]_b^a = \frac{1}{a - b} \left( -\frac{1}{a} - (-\frac{1}{b}) \right)$.
$I = \frac{1}{a - b} \left( \frac{1}{b} - \frac{1}{a} \right) = \frac{1}{a - b} \left( \frac{a - b}{ab} \right) = \frac{1}{ab}$.

(B) Given the integral $\int_0^k \frac{dx}{2 + 8x^2} = \frac{\pi}{16}$.
Factor out $2$ from the denominator: $\frac{1}{2} \int_0^k \frac{dx}{1 + 4x^2} = \frac{1}{2} \int_0^k \frac{dx}{1 + (2x)^2}$.
Let $t = 2x$,then $dt = 2dx$ or $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=k, t=2k$.
The integral becomes $\frac{1}{2} \int_0^{2k} \frac{dt/2}{1 + t^2} = \frac{1}{4} \int_0^{2k} \frac{dt}{1 + t^2}$.
Evaluating the integral: $\frac{1}{4} [\tan^{-1} t]_0^{2k} = \frac{1}{4} \tan^{-1}(2k)$.
Equating to the given value: $\frac{1}{4} \tan^{-1}(2k) = \frac{\pi}{16}$.
$\tan^{-1}(2k) = \frac{\pi}{4}$.
$2k = \tan(\frac{\pi}{4}) = 1$.
Therefore,$k = \frac{1}{2}$.

(B) Let $I = \int_0^{1/\sqrt{2}} \frac{\sin^{-1}x}{(1-x^2)^{3/2}} dx$.
Substitute $\sin^{-1}x = t$,which implies $x = \sin t$ and $dx = \cos t \, dt$.
When $x = 0$,$t = 0$. When $x = 1/\sqrt{2}$,$t = \pi/4$.
The integral becomes $I = \int_0^{\pi/4} \frac{t \cos t}{(1-\sin^2 t)^{3/2}} dt = \int_0^{\pi/4} \frac{t \cos t}{\cos^3 t} dt = \int_0^{\pi/4} t \sec^2 t \, dt$.
Using integration by parts: $\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = \sec^2 t \, dt$.
Then $du = dt$ and $v = \tan t$.
$I = [t \tan t]_0^{\pi/4} - \int_0^{\pi/4} \tan t \, dt$.
$I = [\frac{\pi}{4} \tan(\frac{\pi}{4}) - 0] - [\log|\sec t|]_0^{\pi/4}$.
$I = \frac{\pi}{4}(1) - [\log(\sec \frac{\pi}{4}) - \log(\sec 0)]$.
$I = \frac{\pi}{4} - [\log(\sqrt{2}) - \log(1)] = \frac{\pi}{4} - \log(2^{1/2}) = \frac{\pi}{4} - \frac{1}{2} \log 2$.

(D) Let $I = \int_0^{\pi /2} {\sin x \sin 2x \, dx}$.
Using the trigonometric identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int_0^{\pi /2} {\sin x (2 \sin x \cos x) \, dx} = 2 \int_0^{\pi /2} {\sin^2 x \cos x \, dx}$.
Let $t = \sin x$. Then $dt = \cos x \, dx$.
When $x = 0$,$t = \sin 0 = 0$. When $x = \pi / 2$,$t = \sin(\pi / 2) = 1$.
Substituting these into the integral:
$I = 2 \int_0^1 {t^2 \, dt} = 2 \left[ \frac{t^3}{3} \right]_0^1 = 2 \left( \frac{1}{3} - 0 \right) = \frac{2}{3}$.

(C) Let $I = \int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}}} $.
Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$,we have $2 + \cos x = 2(\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}) + \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \sin^2 \frac{x}{2} + 3\cos^2 \frac{x}{2}$.
Thus,$I = \int_0^{\pi /2} {\frac{{dx}}{{\sin^2 \frac{x}{2} + 3\cos^2 \frac{x}{2}}}} $.
Dividing numerator and denominator by $\cos^2 \frac{x}{2}$,we get $I = \int_0^{\pi /2} {\frac{{\sec^2 \frac{x}{2}}}{{3 + \tan^2 \frac{x}{2}}}} dx$.
Let $t = \tan \frac{x}{2}$,then $dt = \frac{1}{2} \sec^2 \frac{x}{2} dx$,which implies $\sec^2 \frac{x}{2} dx = 2 dt$.
When $x = 0$,$t = 0$. When $x = \pi/2$,$t = \tan(\pi/4) = 1$.
Substituting these,$I = \int_0^1 \frac{2 dt}{3 + t^2} = 2 \int_0^1 \frac{dt}{(\sqrt{3})^2 + t^2}$.
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})$,we get $I = 2 [\frac{1}{\sqrt{3}} \tan^{-1}(\frac{t}{\sqrt{3}})]_0^1 = \frac{2}{\sqrt{3}} \tan^{-1}(\frac{1}{\sqrt{3}})$.
Therefore,the correct option is $C$.

(A) Let $I = \int_0^a \frac{x \, dx}{\sqrt{a^2 + x^2}}$.
Substitute $t = a^2 + x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$.
When $x = 0$,$t = a^2 + 0^2 = a^2$.
When $x = a$,$t = a^2 + a^2 = 2a^2$.
Now,the integral becomes:
$I = \int_{a^2}^{2a^2} \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2} \int_{a^2}^{2a^2} t^{-1/2} \, dt$
$I = \frac{1}{2} [2t^{1/2}]_{a^2}^{2a^2} = [\sqrt{t}]_{a^2}^{2a^2}$
$I = \sqrt{2a^2} - \sqrt{a^2} = a\sqrt{2} - a = a(\sqrt{2} - 1)$.

(A) Let $I = \int_0^a {\frac{{{x^4}\,dx}}{{{{({a^2} + {x^2})}^4}}}} $.
Put $x = a \tan \theta $,then $dx = a \sec^2 \theta \, d\theta $.
When $x = 0$,$\theta = 0$. When $x = a$,$\theta = \frac{\pi}{4}$.
$I = \int_0^{\pi/4} \frac{a^4 \tan^4 \theta \cdot a \sec^2 \theta \, d\theta}{(a^2 + a^2 \tan^2 \theta)^4} = \int_0^{\pi/4} \frac{a^5 \tan^4 \theta \sec^2 \theta \, d\theta}{a^8 \sec^8 \theta} = \frac{1}{a^3} \int_0^{\pi/4} \sin^4 \theta \cos^2 \theta \, d\theta$.
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have $\sin^4 \theta \cos^2 \theta = (\frac{1 - \cos 2\theta}{2})^2 (\frac{1 + \cos 2\theta}{2}) = \frac{1}{8} (1 - 2\cos 2\theta + \cos^2 2\theta)(1 + \cos 2\theta) = \frac{1}{8} (1 - \cos 2\theta - \cos^2 2\theta + \cos^3 2\theta)$.
Using $\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}$ and $\cos^3 2\theta = \frac{3\cos 2\theta + \cos 6\theta}{4}$,
$I = \frac{1}{8a^3} \int_0^{\pi/4} (1 - \cos 2\theta - \frac{1 + \cos 4\theta}{2} + \frac{3\cos 2\theta + \cos 6\theta}{4}) \, d\theta = \frac{1}{32a^3} \int_0^{\pi/4} (4 - 4\cos 2\theta - 2 - 2\cos 4\theta + 3\cos 2\theta + \cos 6\theta) \, d\theta = \frac{1}{32a^3} \int_0^{\pi/4} (2 - \cos 2\theta - 2\cos 4\theta + \cos 6\theta) \, d\theta$.
$I = \frac{1}{32a^3} [2\theta - \frac{\sin 2\theta}{2} - \frac{\sin 4\theta}{2} + \frac{\sin 6\theta}{6}]_0^{\pi/4} = \frac{1}{32a^3} [2(\frac{\pi}{4}) - \frac{1}{2}(1) - \frac{1}{2}(0) + \frac{1}{6}(-1)] = \frac{1}{32a^3} [\frac{\pi}{2} - \frac{1}{2} - \frac{1}{6}] = \frac{1}{32a^3} [\frac{\pi}{2} - \frac{2}{3}] = \frac{1}{16a^3} [\frac{\pi}{4} - \frac{1}{3}]$.

(C) Let $I = \int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$.
Substitute $t = \frac{x}{2}$,then $dt = \frac{1}{2} dx$,which implies $dx = 2 dt$.
When $x = 0$,$t = 0$. When $x = 2\pi$,$t = \pi$.
So,$I = \int_0^{\pi} e^t \sin \left( t + \frac{\pi}{4} \right) (2 dt) = 2 \int_0^{\pi} e^t \sin \left( t + \frac{\pi}{4} \right) dt$.
Using the standard integral formula $\int e^{at} \sin(bt + c) dt = \frac{e^{at}}{a^2 + b^2} [a \sin(bt + c) - b \cos(bt + c)]$,where $a=1, b=1, c=\frac{\pi}{4}$:
$I = 2 \left[ \frac{e^t}{1^2 + 1^2} (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4})) \right]_0^{\pi}$.
$I = 2 \left[ \frac{e^t}{2} (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4})) \right]_0^{\pi} = [e^t (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4}))]_0^{\pi}$.
At $t = \pi$: $e^{\pi} (\sin(\pi + \frac{\pi}{4}) - \cos(\pi + \frac{\pi}{4})) = e^{\pi} (-\sin \frac{\pi}{4} - (-\cos \frac{\pi}{4})) = e^{\pi} (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 0$.
At $t = 0$: $e^0 (\sin \frac{\pi}{4} - \cos \frac{\pi}{4}) = 1 (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$.
Thus,$I = 0 - 0 = 0$.

(D) Let $I = \int_0^1 \frac{e^{-x}}{1 + e^{-x}} \,dx$.
Substitute $t = 1 + e^{-x}$,then $dt = -e^{-x} \,dx$,which implies $e^{-x} \,dx = -dt$.
When $x = 0$,$t = 1 + e^0 = 2$.
When $x = 1$,$t = 1 + e^{-1} = 1 + \frac{1}{e} = \frac{e+1}{e}$.
Substituting these into the integral:
$I = \int_2^{\frac{e+1}{e}} \frac{-dt}{t} = -[\log |t|]_2^{\frac{e+1}{e}}$.
$I = -\left( \log \left( \frac{e+1}{e} \right) - \log 2 \right) = \log 2 - \log \left( \frac{e+1}{e} \right) = \log \left( \frac{2e}{e+1} \right)$.
Wait,let us re-evaluate the original integral:
$I = \int_0^1 \frac{1}{e^x + 1} \,dx = \int_0^1 \frac{e^x}{e^x(e^x+1)} \,dx$.
Let $u = e^x$,$du = e^x \,dx$.
$I = \int_1^e \frac{du}{u(u+1)} = \int_1^e \left( \frac{1}{u} - \frac{1}{u+1} \right) \,du = [\log u - \log(u+1)]_1^e = [\log \frac{u}{u+1}]_1^e = \log \frac{e}{e+1} - \log \frac{1}{2} = \log \frac{2e}{e+1}$.
Since $\log \frac{2e}{e+1} = \log 2 + \log e - \log(e+1) = \log 2 + 1 - \log(e+1)$.
Comparing with options,none match. Thus,the correct answer is $(D)$.

(A) Let $I = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx}$.
Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x)dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi /4$,$t = \sin(\pi /4) - \cos(\pi /4) = 0$.
Also,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - \sin 2x$,so $\sin 2x = 1 - t^2$.
Substituting these into the integral:
$I = \int_{-1}^0 \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^0 \frac{dt}{9 + 16 - 16t^2} = \int_{-1}^0 \frac{dt}{25 - 16t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$,we have:
$I = \frac{1}{16} \int_{-1}^0 \frac{dt}{(5/4)^2 - t^2} = \frac{1}{16} \cdot \frac{1}{2(5/4)} \left[ \log \left| \frac{5/4 + t}{5/4 - t} \right| \right]_{-1}^0$.
$I = \frac{1}{16} \cdot \frac{2}{5} \left[ \log \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^0 = \frac{1}{40} \left[ \log(1) - \log \left| \frac{5 - 4}{5 + 4} \right| \right]$.
$I = \frac{1}{40} [0 - \log(1/9)] = \frac{1}{40} \log(9) = \frac{1}{40} \log(3^2) = \frac{2}{40} \log 3 = \frac{1}{20} \log 3$.

(B) Let $t = \sin^{-1} x$.
Then $dt = \frac{1}{\sqrt{1 - x^2}} dx$.
When $x = 0$,$t = 0$.
When $x = 1/2$,$t = \sin^{-1}(1/2) = \pi/6$.
Also,$x = \sin t$.
Substituting these into the integral:
$\int_0^{\pi/6} t \sin t \, dt$.
Using integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = t$,$dv = \sin t \, dt$.
Then $du = dt$,$v = -\cos t$.
$\int t \sin t \, dt = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t$.
Evaluating the definite integral:
$[-t \cos t + \sin t]_0^{\pi/6} = (-\frac{\pi}{6} \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})) - (0 + 0)$.
$= -\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1}{2} - \frac{\sqrt{3} \pi}{12}$.

(A) Let $I = \int_0^2 \sqrt{\frac{2 + x}{2 - x}} \,dx$.
Put $x = 2 \cos \theta$,so $dx = -2 \sin \theta \,d\theta$.
When $x = 0$,$\cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2}$.
When $x = 2$,$\cos \theta = 1 \Rightarrow \theta = 0$.
Substituting these into the integral:
$I = \int_{\pi/2}^0 \sqrt{\frac{2 + 2 \cos \theta}{2 - 2 \cos \theta}} (-2 \sin \theta) \,d\theta$
$I = 2 \int_0^{\pi/2} \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} \sin \theta \,d\theta$
Using half-angle identities $1 + \cos \theta = 2 \cos^2(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$:
$I = 2 \int_0^{\pi/2} \frac{\cos(\theta/2)}{\sin(\theta/2)} \cdot 2 \sin(\theta/2) \cos(\theta/2) \,d\theta$
$I = 4 \int_0^{\pi/2} \cos^2(\theta/2) \,d\theta$
Using $\cos^2(\theta/2) = \frac{1 + \cos \theta}{2}$:
$I = 4 \int_0^{\pi/2} \frac{1 + \cos \theta}{2} \,d\theta = 2 \int_0^{\pi/2} (1 + \cos \theta) \,d\theta$
$I = 2 [\theta + \sin \theta]_0^{\pi/2} = 2 [(\frac{\pi}{2} + 1) - (0 + 0)] = \pi + 2$.

(C) We evaluate the integral $I = \int_0^\pi \frac{dx}{1 + \sin x}$.
Multiply the numerator and denominator by $(1 - \sin x)$:
$I = \int_0^\pi \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx = \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x} dx = \int_0^\pi \frac{1 - \sin x}{\cos^2 x} dx$.
Split the integral into two parts:
$I = \int_0^\pi (\sec^2 x - \sec x \tan x) dx$.
Now,integrate each term:
$I = [\tan x - \sec x]_0^\pi$.
Evaluate at the limits:
$I = (\tan \pi - \sec \pi) - (\tan 0 - \sec 0)$.
Since $\tan \pi = 0$,$\sec \pi = -1$,$\tan 0 = 0$,and $\sec 0 = 1$:
$I = (0 - (-1)) - (0 - 1) = 1 - (-1) = 1 + 1 = 2$.

(A) We are given the integral $I = \int_0^{\pi /8} \frac{\sec^2 2x}{2} \, dx$.
Taking the constant factor out,we get $I = \frac{1}{2} \int_0^{\pi /8} \sec^2 2x \, dx$.
We know that the integral of $\sec^2(ax)$ is $\frac{\tan(ax)}{a}$.
Applying this,we get $I = \frac{1}{2} \left[ \frac{\tan 2x}{2} \right]_0^{\pi /8}$.
$I = \frac{1}{4} [\tan 2x]_0^{\pi /8}$.
Substituting the limits,$I = \frac{1}{4} [\tan(2 \times \frac{\pi}{8}) - \tan(0)]$.
$I = \frac{1}{4} [\tan(\frac{\pi}{4}) - 0]$.
Since $\tan(\frac{\pi}{4}) = 1$,we have $I = \frac{1}{4} [1] = \frac{1}{4}$.

(C) We know that $1 + \sin \theta = (\sin \frac{\theta}{2} + \cos \frac{\theta}{2})^2$.
Thus,$\sqrt{1 + \sin \frac{x}{2}} = \sqrt{(\sin \frac{x}{4} + \cos \frac{x}{4})^2} = |\sin \frac{x}{4} + \cos \frac{x}{4}|$.
Since $x \in [0, 2\pi]$,$\frac{x}{4} \in [0, \frac{\pi}{2}]$,where both $\sin \frac{x}{4}$ and $\cos \frac{x}{4}$ are non-negative.
Therefore,the integral becomes $\int_0^{2\pi} (\sin \frac{x}{4} + \cos \frac{x}{4}) dx$.
Evaluating the integral:
$= [-4 \cos \frac{x}{4} + 4 \sin \frac{x}{4}]_0^{2\pi}$
$= (-4 \cos \frac{\pi}{2} + 4 \sin \frac{\pi}{2}) - (-4 \cos 0 + 4 \sin 0)$
$= (-4(0) + 4(1)) - (-4(1) + 4(0))$
$= 4 - (-4) = 8$.

(B) To evaluate the integral $I = \int_0^1 {{\cos }^{ - 1}}x\,dx$,we use integration by parts: $\int u\,dv = uv - \int v\,du$.
Let $u = {\cos }^{ - 1}x$ and $dv = dx$.
Then $du = -\frac{1}{\sqrt{1 - x^2}}dx$ and $v = x$.
Applying the formula: $I = [x{\cos }^{ - 1}x]_0^1 - \int_0^1 x \left( -\frac{1}{\sqrt{1 - x^2}} \right) dx$.
$I = [x{\cos }^{ - 1}x]_0^1 + \int_0^1 \frac{x}{\sqrt{1 - x^2}} dx$.
For the second integral,let $t = 1 - x^2$,so $dt = -2x\,dx$ or $x\,dx = -\frac{1}{2}dt$.
When $x=0, t=1$; when $x=1, t=0$.
$I = [x{\cos }^{ - 1}x]_0^1 - \frac{1}{2} \int_1^0 \frac{1}{\sqrt{t}} dt = [x{\cos }^{ - 1}x]_0^1 - \frac{1}{2} [2\sqrt{t}]_1^0$.
$I = (1 \cdot {\cos }^{ - 1}(1) - 0 \cdot {\cos }^{ - 1}(0)) - (\sqrt{0} - \sqrt{1}) = (0 - 0) - (0 - 1) = 1$.

(D) Let $I = \int_0^{\pi /6} {(2 + 3{x^2})\cos 3x\,dx}$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = 2 + 3x^2$ and $dv = \cos 3x \, dx$.
Then $du = 6x \, dx$ and $v = \frac{\sin 3x}{3}$.
$I = \left[ (2 + 3x^2) \frac{\sin 3x}{3} \right]_0^{\pi /6} - \int_0^{\pi /6} \frac{\sin 3x}{3} (6x) \, dx$
$I = \left[ (2 + 3(\frac{\pi^2}{36})) \frac{\sin(\pi/2)}{3} - 0 \right] - 2 \int_0^{\pi /6} x \sin 3x \, dx$
$I = \frac{1}{3} (2 + \frac{\pi^2}{12}) - 2 \left[ x \left( -\frac{\cos 3x}{3} \right) - \int_0^{\pi /6} (1) \left( -\frac{\cos 3x}{3} \right) \, dx \right]$
$I = \frac{2}{3} + \frac{\pi^2}{36} - 2 \left[ -\frac{x \cos 3x}{3} + \frac{1}{3} \int_0^{\pi /6} \cos 3x \, dx \right]$
$I = \frac{2}{3} + \frac{\pi^2}{36} + \frac{2x \cos 3x}{3} - \frac{2}{3} \left[ \frac{\sin 3x}{3} \right]_0^{\pi /6}$
Evaluating at limits: $I = \frac{2}{3} + \frac{\pi^2}{36} + \frac{2}{3} (\frac{\pi}{6} \cos(\pi/2)) - \frac{2}{9} (\sin(\pi/2) - \sin 0)$
Since $\cos(\pi/2) = 0$ and $\sin(\pi/2) = 1$,
$I = \frac{2}{3} + \frac{\pi^2}{36} + 0 - \frac{2}{9} = \frac{6-2}{9} + \frac{\pi^2}{36} = \frac{4}{9} + \frac{\pi^2}{36} = \frac{16 + \pi^2}{36} = \frac{1}{36}(\pi^2 + 16)$.

(D) Let $t = x^2 + 1$,then $dt = 2x \, dx$ or $x \, dx = \frac{1}{2} \, dt$.
When $x = 0$,$t = 1$. When $x = 2$,$t = 5$.
The integral becomes $\int_0^2 \frac{x^2 \cdot x \, dx}{(x^2 + 1)^{3/2}} = \frac{1}{2} \int_1^5 \frac{t - 1}{t^{3/2}} \, dt$.
$= \frac{1}{2} \int_1^5 (t^{-1/2} - t^{-3/2}) \, dt$.
$= \frac{1}{2} [2t^{1/2} - \frac{t^{-1/2}}{-1/2}]_1^5 = \frac{1}{2} [2\sqrt{t} + \frac{2}{\sqrt{t}}]_1^5$.
$= [\sqrt{t} + \frac{1}{\sqrt{t}}]_1^5 = (\sqrt{5} + \frac{1}{\sqrt{5}}) - (1 + 1) = \frac{5+1}{\sqrt{5}} - 2 = \frac{6}{\sqrt{5}} - 2 = \frac{6 - 2\sqrt{5}}{\sqrt{5}}$.
Since this value is not among the options,the correct choice is $D$.

(B) Let $I = \int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} $.
Substitute $\cos x = t$,then $-\sin x\,dx = dt$.
When $x = 0$,$t = 1$. When $x = \pi/2$,$t = 0$.
Thus,$I = \int_1^0 {\frac{-t\,dt}{{{t^2} + 3t + 2}}} = \int_0^1 {\frac{t\,dt}{{(t+1)(t+2)}}} $.
Using partial fractions,$\frac{t}{(t+1)(t+2)} = \frac{2}{t+2} - \frac{1}{t+1}$.
$I = \int_0^1 {\left( {\frac{2}{{t + 2}} - \frac{1}{{t + 1}}} \right)} \,dt$.
$I = [2\log |t + 2| - \log |t + 1|]_0^1$.
$I = (2\log 3 - \log 2) - (2\log 2 - \log 1)$.
Since $\log 1 = 0$,$I = 2\log 3 - 3\log 2 = \log 3^2 - \log 2^3 = \log 9 - \log 8 = \log \left( {\frac{9}{8}} \right)$.

(D) We have the integral $I = \int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$.
Completing the square in the denominator: $x^2 + 2x\cos \alpha + \cos^2 \alpha + 1 - \cos^2 \alpha = (x + \cos \alpha)^2 + \sin^2 \alpha$.
Thus,$I = \int_0^1 \frac{dx}{(x + \cos \alpha)^2 + \sin^2 \alpha}$.
Using the standard integral $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \left[ \frac{1}{\sin \alpha} \tan^{-1}\left( \frac{x + \cos \alpha}{\sin \alpha} \right) \right]_0^1$.
Evaluating at the limits:
$I = \frac{1}{\sin \alpha} \left[ \tan^{-1}\left( \frac{1 + \cos \alpha}{\sin \alpha} \right) - \tan^{-1}\left( \frac{\cos \alpha}{\sin \alpha} \right) \right]$.
Using trigonometric identities $\frac{1 + \cos \alpha}{\sin \alpha} = \cot(\frac{\alpha}{2})$ and $\frac{\cos \alpha}{\sin \alpha} = \cot \alpha$:
$I = \frac{1}{\sin \alpha} \left[ \tan^{-1}(\cot \frac{\alpha}{2}) - \tan^{-1}(\cot \alpha) \right]$.
Since $\tan^{-1}(\cot \theta) = \frac{\pi}{2} - \theta$:
$I = \frac{1}{\sin \alpha} \left[ (\frac{\pi}{2} - \frac{\alpha}{2}) - (\frac{\pi}{2} - \alpha) \right] = \frac{1}{\sin \alpha} [\frac{\alpha}{2}] = \frac{\alpha}{2 \sin \alpha}$.

(A) We use the standard integral formula: $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C$.
Here,$a = -1$ and $b = 1$.
Thus,$\int e^{-x} \sin x dx = \frac{e^{-x}}{(-1)^2 + 1^2} (-1 \sin x - 1 \cos x) = -\frac{e^{-x}}{2} (\sin x + \cos x)$.
Now,applying the limits from $-\pi/4$ to $\pi/2$:
$= [-\frac{e^{-x}}{2} (\sin x + \cos x)]_{-\pi/4}^{\pi/2}$
$= -\frac{1}{2} [e^{-\pi/2} (\sin(\pi/2) + \cos(\pi/2)) - e^{\pi/4} (\sin(-\pi/4) + \cos(-\pi/4))]$
$= -\frac{1}{2} [e^{-\pi/2} (1 + 0) - e^{\pi/4} (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})]$
$= -\frac{1}{2} [e^{-\pi/2} (1) - e^{\pi/4} (0)]$
$= -\frac{1}{2} e^{-\pi/2}$.

(C) Let $I = \int_0^{\pi /2} \frac{1 + 2\cos x}{(2 + \cos x)^2} dx$.
We can rewrite the numerator as $2(2 + \cos x) - 3$.
So,$I = \int_0^{\pi /2} \frac{2(2 + \cos x) - 3}{(2 + \cos x)^2} dx = 2 \int_0^{\pi /2} \frac{1}{2 + \cos x} dx - 3 \int_0^{\pi /2} \frac{1}{(2 + \cos x)^2} dx$.
Using the substitution $t = \tan(x/2)$,we have $\cos x = \frac{1 - t^2}{1 + t^2}$ and $dx = \frac{2 dt}{1 + t^2}$.
As $x$ goes from $0$ to $\pi/2$,$t$ goes from $0$ to $1$.
$I = 2 \int_0^1 \frac{1}{2 + \frac{1 - t^2}{1 + t^2}} \cdot \frac{2 dt}{1 + t^2} - 3 \int_0^1 \frac{1}{(2 + \frac{1 - t^2}{1 + t^2})^2} \cdot \frac{2 dt}{1 + t^2} = 4 \int_0^1 \frac{dt}{3 + t^2} - 6 \int_0^1 \frac{1 + t^2}{(3 + t^2)^2} dt$.
Using integration by parts on the second integral or reduction formulas,we find the result is $\frac{1}{2}$.

(C) Let $I = \int_0^\pi \frac{dx}{1 - 2a\cos x + a^2}$.
Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$,we get:
$I = \int_0^\pi \frac{dx}{(1+a^2)(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) - 2a(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$
$I = \int_0^\pi \frac{dx}{(1-a)^2 \cos^2 \frac{x}{2} + (1+a)^2 \sin^2 \frac{x}{2}}$
Divide numerator and denominator by $\cos^2 \frac{x}{2}$:
$I = \int_0^\pi \frac{\sec^2 \frac{x}{2} dx}{(1-a)^2 + (1+a)^2 \tan^2 \frac{x}{2}}$
Let $t = \tan \frac{x}{2}$,then $dt = \frac{1}{2} \sec^2 \frac{x}{2} dx$,so $\sec^2 \frac{x}{2} dx = 2 dt$.
As $x \to 0, t \to 0$ and as $x \to \pi, t \to \infty$.
$I = \int_0^\infty \frac{2 dt}{(1-a)^2 + (1+a)^2 t^2} = \frac{2}{(1+a)^2} \int_0^\infty \frac{dt}{(\frac{1-a}{1+a})^2 + t^2}$
Using $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} \tan^{-1}(\frac{x}{k})$:
$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} [\tan^{-1}(\frac{1+a}{1-a} t)]_0^\infty = \frac{2}{1-a^2} [\frac{\pi}{2} - 0] = \frac{\pi}{1-a^2}$.

(B) Let $I = \int_0^1 {(1 - x)^9} dx$.
Using the substitution method,let $u = 1 - x$,then $du = -dx$,which implies $dx = -du$.
When $x = 0$,$u = 1$.
When $x = 1$,$u = 0$.
Substituting these into the integral:
$I = \int_1^0 u^9 (-du) = \int_0^1 u^9 du$.
Evaluating the integral:
$I = \left[ \frac{u^{10}}{10} \right]_0^1 = \frac{1^{10}}{10} - \frac{0^{10}}{10} = \frac{1}{10} - 0 = \frac{1}{10}$.

(A) To evaluate the definite integral $\int_0^{\pi /3} \cos 3x \, dx$,we first find the antiderivative of $\cos 3x$.
The integral of $\cos 3x$ is $\frac{\sin 3x}{3}$.
Applying the fundamental theorem of calculus:
$\int_0^{\pi /3} \cos 3x \, dx = \left[ \frac{\sin 3x}{3} \right]_0^{\pi /3}$
$= \frac{\sin(3 \times \frac{\pi}{3})}{3} - \frac{\sin(3 \times 0)}{3}$
$= \frac{\sin(\pi)}{3} - \frac{\sin(0)}{3}$
$= \frac{0}{3} - \frac{0}{3} = 0$.
Thus,the correct option is $A$.

(D) We know that $\frac{1 + \tan x}{1 - \tan x} = \tan(\frac{\pi}{4} + x)$.
Therefore,the integral becomes $I = \int_0^{\pi/4} \tan(\frac{\pi}{4} + x) \, dx$.
Using the formula $\int \tan(ax+b) \, dx = \frac{1}{a} \ln|\sec(ax+b)| + C$,we get:
$I = [\ln|\sec(\frac{\pi}{4} + x)|]_0^{\pi/4}$.
Evaluating at the limits:
$I = \ln|\sec(\frac{\pi}{4} + \frac{\pi}{4})| - \ln|\sec(\frac{\pi}{4} + 0)|$.
$I = \ln|\sec(\frac{\pi}{2})| - \ln|\sec(\frac{\pi}{4})|$.
Since $\sec(\frac{\pi}{2})$ is undefined,let us re-evaluate using $\frac{1 + \tan x}{1 - \tan x} = \tan(\frac{\pi}{4} + x)$.
Actually,the integral $\int_0^{\pi/4} \tan(\frac{\pi}{4} + x) \, dx = [\ln|\sec(\frac{\pi}{4} + x)|]_0^{\pi/4} = \ln(\sec \frac{\pi}{2}) - \ln(\sec \frac{\pi}{4})$.
Since $\sec(\frac{\pi}{2}) \to \infty$,the integral diverges. Thus,the correct option is $D$.

(B) We have the integral $I = \int_0^1 \frac{dx}{e^x + e^{-x}}$.
Multiply the numerator and denominator by $e^x$:
$I = \int_0^1 \frac{e^x}{e^{2x} + 1} dx$.
Let $t = e^x$,then $dt = e^x dx$.
When $x = 0$,$t = e^0 = 1$. When $x = 1$,$t = e^1 = e$.
Substituting these into the integral:
$I = \int_1^e \frac{dt}{1 + t^2} = [\tan^{-1} t]_1^e$.
Evaluating the limits:
$I = \tan^{-1}(e) - \tan^{-1}(1) = \tan^{-1}(e) - \frac{\pi}{4}$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x - y}{1 + xy}\right)$:
$I = \tan^{-1}\left(\frac{e - 1}{1 + e \cdot 1}\right) = \tan^{-1}\left(\frac{e - 1}{e + 1}\right)$.

(C) To evaluate the integral $I = \int_0^1 x \log \left( 1 + \frac{x}{2} \right) dx$,we use integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = \log \left( 1 + \frac{x}{2} \right)$ and $v = x$.
Then $du = \frac{1}{1 + x/2} \cdot \frac{1}{2} dx = \frac{1}{x+2} dx$ and $\int v dx = \frac{x^2}{2}$.
$I = \left[ \frac{x^2}{2} \log \left( 1 + \frac{x}{2} \right) \right]_0^1 - \int_0^1 \frac{x^2}{2(x+2)} dx$
$I = \left( \frac{1}{2} \log \frac{3}{2} - 0 \right) - \frac{1}{2} \int_0^1 \frac{x^2}{x+2} dx$
Using polynomial division,$\frac{x^2}{x+2} = x - 2 + \frac{4}{x+2}$.
$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \int_0^1 \left( x - 2 + \frac{4}{x+2} \right) dx$
$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \left[ \frac{x^2}{2} - 2x + 4 \log(x+2) \right]_0^1$
$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \left( \left( \frac{1}{2} - 2 + 4 \log 3 \right) - (0 - 0 + 4 \log 2) \right)$
$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \left( -\frac{3}{2} + 4 \log \frac{3}{2} \right)$
$I = \frac{1}{2} \log \frac{3}{2} + \frac{3}{4} - 2 \log \frac{3}{2} = \frac{3}{4} - \frac{3}{2} \log \frac{3}{2} = \frac{3}{4} + \frac{3}{2} \log \frac{2}{3}$.
Comparing with $a + b \log \frac{2}{3}$,we get $a = \frac{3}{4}$ and $b = \frac{3}{2}$.

(B) Let $I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$.
Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+x} + \sqrt{x})$:
$I = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x} dx = \int_0^1 (\sqrt{1+x} + \sqrt{x}) dx$.
Integrating term by term:
$I = \left[ \frac{2}{3}(1+x)^{3/2} + \frac{2}{3}x^{3/2} \right]_0^1$.
Evaluating at the limits:
$I = \left( \frac{2}{3}(2)^{3/2} + \frac{2}{3}(1)^{3/2} \right) - \left( \frac{2}{3}(1)^{3/2} + 0 \right)$.
$I = \frac{2}{3}(2\sqrt{2}) + \frac{2}{3} - \frac{2}{3} = \frac{4\sqrt{2}}{3}$.

(C) Let $I = \int_0^{\pi /4} {\frac{{4\sin 2\theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }}} = \int_0^{\pi /4} {\frac{{8\sin \theta \cos \theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }}} $
Dividing the numerator and denominator by $\cos^4 \theta$,we get:
$I = \int_0^{\pi /4} {\frac{{8\tan \theta \sec^2 \theta \,d\theta }}{{{\tan^4 \theta + 1}}} } $
Let $\tan^2 \theta = t$. Then $2 \tan \theta \sec^2 \theta \,d\theta = dt$.
When $\theta = 0$,$t = 0$. When $\theta = \pi/4$,$t = 1$.
Substituting these into the integral:
$I = 4 \int_0^1 {\frac{{dt}}{{{t^2} + 1}}} $
$I = 4 [\tan^{-1} t]_0^1 = 4(\frac{\pi}{4} - 0) = \pi $.

(A) Given $x({x^4} + 1)\phi (x) = 1,$ we have $\phi (x) = \frac{1}{{x({x^4} + 1)}}.$
We can write $\phi (x)$ using partial fractions as:
$\phi (x) = \frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}.$
Now,we integrate $\phi (x)$ from $1$ to $2$:
$\int_1^2 {\phi (x)\,dx = \int_1^2 {\left( {\frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}} \right)\,dx} }.$
$= \left[ \log |x| \right]_1^2 - \int_1^2 {\frac{{{x^3}}}{{{x^4} + 1}}\,dx}.$
Let $u = {x^4} + 1,$ then $du = 4{x^3}\,dx,$ so ${x^3}\,dx = \frac{1}{4}du.$
When $x = 1, u = 2.$ When $x = 2, u = 17.$
$= (\log 2 - \log 1) - \frac{1}{4} \int_2^{17} {\frac{1}{u}\,du} = \log 2 - \frac{1}{4} \left[ \log u \right]_2^{17}.$
$= \log 2 - \frac{1}{4} (\log 17 - \log 2) = \log 2 - \frac{1}{4} \log 17 + \frac{1}{4} \log 2.$
$= \frac{5}{4} \log 2 - \frac{1}{4} \log 17 = \frac{1}{4} (5 \log 2 - \log 17) = \frac{1}{4} \log \left( \frac{2^5}{17} \right) = \frac{1}{4} \log \frac{32}{17}.$
Thus,the correct option is $A$.

(D) Let $I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}}$.
We can rewrite the expression inside the square root by completing the square:
$x - x^2 = -(x^2 - x) = -((x - 1/2)^2 - 1/4) = (1/2)^2 - (x - 1/2)^2$.
Thus,$I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{(1/2)^2 - (x - 1/2)^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,we get:
$I = [\sin^{-1}(\frac{x - 1/2}{1/2})]_{1/4}^{1/2} = [\sin^{-1}(2x - 1)]_{1/4}^{1/2}$.
Evaluating at the limits:
$I = \sin^{-1}(2(1/2) - 1) - \sin^{-1}(2(1/4) - 1) = \sin^{-1}(0) - \sin^{-1}(-1/2)$.
$I = 0 - (-\pi/6) = \pi/6$.

(A) We need to evaluate the definite integral $I = \int_0^{2\pi} (\sin x + \cos x) \, dx$.
Applying the fundamental theorem of calculus,we find the antiderivative of $(\sin x + \cos x)$,which is $(-\cos x + \sin x)$.
Evaluating this at the limits $0$ and $2\pi$:
$I = [-\cos x + \sin x]_0^{2\pi}$
$I = (-\cos(2\pi) + \sin(2\pi)) - (-\cos(0) + \sin(0))$
Since $\cos(2\pi) = 1$,$\sin(2\pi) = 0$,$\cos(0) = 1$,and $\sin(0) = 0$:
$I = (-1 + 0) - (-1 + 0)$
$I = -1 + 1 = 0$
Thus,the correct option is $A$.

(A) Let $I = \int_0^{\pi /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} dx = \int_0^{\pi /4} {\frac{{\cos x}}{{{{\cos }^2}x(1 + 2{{\sin }^2}x)}}} dx$
$= \int_0^{\pi /4} {\frac{{\cos x}}{{(1 - {{\sin }^2}x)(1 + 2{{\sin }^2}x)}}} dx$
Let $t = \sin x$,then $dt = \cos x dx$. When $x=0, t=0$ and when $x=\pi/4, t=1/\sqrt{2}$.
$I = \int_0^{1/\sqrt{2}} \frac{dt}{(1-t^2)(1+2t^2)}$.
Using partial fractions: $\frac{1}{(1-t^2)(1+2t^2)} = \frac{1}{3} \left( \frac{1}{1-t^2} + \frac{2}{1+2t^2} \right)$.
$I = \frac{1}{3} \left[ \int_0^{1/\sqrt{2}} \frac{dt}{1-t^2} + 2 \int_0^{1/\sqrt{2}} \frac{dt}{1+2t^2} \right]$
$I = \frac{1}{3} \left[ \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1}(t\sqrt{2}) \right]_0^{1/\sqrt{2}}$
$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{1+1/\sqrt{2}}{1-1/\sqrt{2}} \right) + \sqrt{2} \tan^{-1}(1) \right]$
$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) + \sqrt{2} \cdot \frac{\pi}{4} \right]$
Since $\frac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^2$,we have $\frac{1}{2} \log (\sqrt{2}+1)^2 = \log(\sqrt{2}+1)$.
$I = \frac{1}{3} \left[ \log(\sqrt{2}+1) + \frac{\pi}{2\sqrt{2}} \right]$.

(C) To evaluate the integral $\int_1^2 \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \log x$ and $dv = dx$. Then $du = \frac{1}{x} dx$ and $v = x$.
Applying the formula:
$\int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx = x \log x - \int 1 \, dx = x \log x - x$.
Now,applying the limits from $1$ to $2$:
$[x \log x - x]_1^2 = (2 \log 2 - 2) - (1 \log 1 - 1)$.
Since $\log 1 = 0$,we have:
$(2 \log 2 - 2) - (0 - 1) = 2 \log 2 - 2 + 1 = 2 \log 2 - 1$.
Using the property $n \log a = \log(a^n)$,we get $2 \log 2 = \log(2^2) = \log 4$.
Also,$1 = \log e$.
Therefore,$\log 4 - \log e = \log(4/e)$.

(B) Given integral $I = \int_3^5 \frac{x^2}{x^2 - 4} \, dx$.
We can rewrite the integrand as:
$\frac{x^2}{x^2 - 4} = \frac{x^2 - 4 + 4}{x^2 - 4} = 1 + \frac{4}{x^2 - 4}$.
Now,$I = \int_3^5 \left( 1 + \frac{4}{x^2 - 4} \right) \, dx$.
Using the partial fraction decomposition $\frac{4}{x^2 - 4} = \frac{4}{(x-2)(x+2)} = \frac{1}{x-2} - \frac{1}{x+2}$.
So,$I = \int_3^5 \left( 1 + \frac{1}{x-2} - \frac{1}{x+2} \right) \, dx$.
Integrating term by term:
$I = [x + \log_e |x-2| - \log_e |x+2|]_3^5$.
$I = [x + \log_e \left| \frac{x-2}{x+2} \right|]_3^5$.
Evaluating at the limits:
$I = (5 + \log_e \frac{3}{7}) - (3 + \log_e \frac{1}{5})$.
$I = 2 + \log_e \left( \frac{3/7}{1/5} \right) = 2 + \log_e \left( \frac{15}{7} \right)$.
Thus,the correct option is $B$.

(A) Let $I = \int_0^{\pi /4} \frac{dx}{\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x}$.
Divide the numerator and denominator by $\cos^4 x$:
$I = \int_0^{\pi /4} \frac{\sec^4 x dx}{1 - \tan^2 x + \tan^4 x} = \int_0^{\pi /4} \frac{(1 + \tan^2 x) \sec^2 x dx}{1 - \tan^2 x + \tan^4 x}$.
Substitute $\tan x = t$,so $\sec^2 x dx = dt$. When $x = 0, t = 0$ and when $x = \pi/4, t = 1$.
$I = \int_0^1 \frac{1 + t^2}{t^4 - t^2 + 1} dt = \int_0^1 \frac{\frac{1}{t^2} + 1}{t^2 - 1 + \frac{1}{t^2}} dt = \int_0^1 \frac{1 + \frac{1}{t^2}}{(t - \frac{1}{t})^2 + 1} dt$.
Let $u = t - \frac{1}{t}$,then $du = (1 + \frac{1}{t^2}) dt$. As $t \to 0^+, u \to -\infty$ and as $t \to 1, u \to 0$.
$I = \int_{-\infty}^0 \frac{du}{u^2 + 1} = [\tan^{-1} u]_{-\infty}^0 = \tan^{-1}(0) - \tan^{-1}(-\infty) = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2}$.