Fundamental definite integration Questions in English

(D) Given the function $f(x) = \int_0^x \frac{\sin t}{t} dt$.
By the Fundamental Theorem of Calculus,the derivative is $f'(x) = \frac{\sin x}{x}$.
For extrema,we set $f'(x) = 0$.
$\frac{\sin x}{x} = 0$ implies $\sin x = 0$ for $x > 0$.
This occurs at $x = n\pi$ for $n = 1, 2, 3, \dots$.
Thus,the points of extrema are $x = n\pi$ where $n \in \mathbb{N}$.

(A) Let $I = \int_2^e {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx$.
Using integration by parts on the first term $\int \frac{1}{\log x} dx$:
Let $u = \frac{1}{\log x}$ and $dv = dx$.
Then $du = -\frac{1}{x(\log x)^2} dx$ and $v = x$.
So,$\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{x(\log x)^2} \right) dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$.
Substituting this back into the integral $I$:
$I = \left[ \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right]_2^e - \int_2^e \frac{1}{(\log x)^2} dx$.
$I = \left[ \frac{x}{\log x} \right]_2^e = \frac{e}{\log e} - \frac{2}{\log 2} = e - \frac{2}{\log 2}$.
Comparing this with $\alpha + \frac{\beta}{\log 2}$,we get $\alpha = e$ and $\beta = -2$.

(A) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.
For the interval $[-1, 1]$,we split the integral at $x = 0$:
$\int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx$.
For $-1 \le x < 0$,$[x] = -1$,so $f(x) = x - (-1) = x + 1$.
For $0 \le x < 1$,$[x] = 0$,so $f(x) = x - 0 = x$.
Now,calculate the integrals:
$\int_{-1}^{0} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^{0} = (0 + 0) - \left( \frac{1}{2} - 1 \right) = -(-1/2) = 1/2$.
$\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = 1/2$.
Adding these results: $1/2 + 1/2 = 1$.

(C) Let $I = \int_{-1/2}^{1/2} \left[ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right]^{1/2} dx$.
Notice that the expression inside the square root is of the form $a^2 + b^2 - 2 = (a-b)^2$,where $a = \frac{x+1}{x-1}$ and $b = \frac{x-1}{x+1}$.
Thus,$\left( \frac{x+1}{x-1} - \frac{x-1}{x+1} \right)^2 = \left( \frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} \right)^2 = \left( \frac{4x}{x^2-1} \right)^2$.
So,$I = \int_{-1/2}^{1/2} \left| \frac{4x}{x^2-1} \right| dx = \int_{-1/2}^{1/2} \left| \frac{4x}{1-x^2} \right| dx$.
Since the integrand is an even function,$I = 2 \int_{0}^{1/2} \frac{4x}{1-x^2} dx = 8 \int_{0}^{1/2} \frac{x}{1-x^2} dx$.
Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.
When $x=0, u=1$; when $x=1/2, u=3/4$.
$I = 8 \int_{1}^{3/4} \frac{-1/2}{u} du = 4 \int_{3/4}^{1} \frac{1}{u} du = 4 [\log |u|]_{3/4}^{1} = 4 (\log 1 - \log(3/4)) = -4 \log(3/4) = 4 \log(4/3)$.

(B) Given $F(t) = \int_{0}^{t} f(t - y) g(y) dy$.
Substituting the given functions $f(y) = e^y$ and $g(y) = y$:
$F(t) = \int_{0}^{t} e^{t - y} y dy = e^t \int_{0}^{t} y e^{-y} dy$.
Using integration by parts for $\int y e^{-y} dy$:
Let $u = y$ and $dv = e^{-y} dy$. Then $du = dy$ and $v = -e^{-y}$.
$\int y e^{-y} dy = -y e^{-y} - \int (-e^{-y}) dy = -y e^{-y} - e^{-y}$.
Evaluating the definite integral from $0$ to $t$:
$F(t) = e^t [ -y e^{-y} - e^{-y} ]_{0}^{t} = e^t [ (-t e^{-t} - e^{-t}) - (0 - 1) ]$.
$F(t) = e^t [ -t e^{-t} - e^{-t} + 1 ] = -t - 1 + e^t$.
Thus,$F(t) = e^t - (1 + t)$.

(A) Let $I = \int_{1}^{\infty} \frac{dx}{e \cdot e^x + e^3 \cdot e^{-x}}$.
Multiply the numerator and denominator by $e^x$:
$I = \int_{1}^{\infty} \frac{e^x \, dx}{e(e^{2x} + e^2)}$.
Substitute $e^x = t$,so $e^x \, dx = dt$. When $x=1, t=e$; when $x \to \infty, t \to \infty$.
$I = \frac{1}{e} \int_{e}^{\infty} \frac{dt}{t^2 + e^2}$.
Using the formula $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{t}{a}$:
$I = \frac{1}{e} \left[ \frac{1}{e} \tan^{-1} \frac{t}{e} \right]_{e}^{\infty} = \frac{1}{e^2} \left[ \tan^{-1}(\infty) - \tan^{-1}(1) \right]$.
$I = \frac{1}{e^2} \left( \frac{\pi}{2} - \frac{\pi}{4} \right) = \frac{\pi}{4e^2}$.

(A) Let $I = \int_{a-c}^{b-c} f(x+c) \, dx$.
Substitute $x + c = t$.
Then $dx = dt$.
When $x = a - c$,$t = (a - c) + c = a$.
When $x = b - c$,$t = (b - c) + c = b$.
Substituting these into the integral,we get:
$I = \int_{a}^{b} f(t) \, dt$.
Since the variable of integration is a dummy variable,we can write $f(t) \, dt$ as $f(x) \, dx$.
Therefore,$I = \int_{a}^{b} f(x) \, dx$.

(C) Let $I = \int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$.
Consider the integral $\int 3x^2 \sin \frac{1}{x} dx$. Using integration by parts,let $u = \sin \frac{1}{x}$ and $dv = 3x^2 dx$. Then $du = -\cos \frac{1}{x} \cdot (-\frac{1}{x^2}) dx = \frac{1}{x^2} \cos \frac{1}{x} dx$ and $v = x^3$.
Thus,$\int 3x^2 \sin \frac{1}{x} dx = x^3 \sin \frac{1}{x} - \int x^3 \cdot \frac{1}{x^2} \cos \frac{1}{x} dx = x^3 \sin \frac{1}{x} - \int x \cos \frac{1}{x} dx$.
Substituting this into the original integral:
$I = \left[ x^3 \sin \frac{1}{x} - \int x \cos \frac{1}{x} dx \right]_{0}^{4/\pi} - \int_{0}^{4/\pi} (-x \cos \frac{1}{x}) dx$.
$I = \left[ x^3 \sin \frac{1}{x} \right]_{0}^{4/\pi} - \int_{0}^{4/\pi} x \cos \frac{1}{x} dx + \int_{0}^{4/\pi} x \cos \frac{1}{x} dx$.
$I = \left[ x^3 \sin \frac{1}{x} \right]_{0}^{4/\pi} = (4/\pi)^3 \sin(\pi/4) - 0 = \frac{64}{\pi^3} \cdot \frac{1}{\sqrt{2}} = \frac{64}{\sqrt{2} \pi^3} = \frac{32 \cdot 2}{\sqrt{2} \pi^3} = \frac{32\sqrt{2}}{\pi^3}$.

(A) Given that $F(x)$ is an antiderivative of $f(x) = \frac{\sin x}{x}$,we have $\int \frac{\sin x}{x} dx = F(x) + C$.
We need to evaluate the integral $I = \int_{1}^{3} \frac{\sin 2x}{x} dx$.
Let $t = 2x$,then $dt = 2 dx$,which implies $dx = \frac{dt}{2}$.
When $x = 1$,$t = 2$. When $x = 3$,$t = 6$.
Substituting these into the integral:
$I = \int_{2}^{6} \frac{\sin t}{t/2} \cdot \frac{dt}{2} = \int_{2}^{6} \frac{\sin t}{t} dt$.
Since $\int \frac{\sin t}{t} dt = F(t)$,by the Fundamental Theorem of Calculus,$\int_{2}^{6} \frac{\sin t}{t} dt = F(6) - F(2)$.
Thus,the correct option is $A$.

(A) Let $I = \int\limits_2^4 {\left[ {{{\log }_x}2 - \frac{{{{\left( {{{\log }_x}2} \right)}^2}}}{{\ln 2}}} \right]} dx$.
Using the change of base formula,$\log_x 2 = \frac{\ln 2}{\ln x}$.
Substituting this into the integral:
$I = \int\limits_2^4 {\left( {\frac{{\ln 2}}{{\ln x}} - \frac{{{{(\ln 2)}^2}}}{{{{(\ln x)}^2} \cdot \ln 2}}} \right)} dx = \ln 2 \int\limits_2^4 {\left( {\frac{1}{{\ln x}} - \frac{1}{{{{(\ln x)}^2}}}} \right)} dx$.
Let $f(x) = \frac{x}{\ln x}$. Then $f'(x) = \frac{\ln x - x(1/x)}{{(\ln x)^2}} = \frac{\ln x - 1}{{(\ln x)^2}} = \frac{1}{\ln x} - \frac{1}{{(\ln x)^2}}$.
Thus,$I = \ln 2 \left[ {\frac{x}{{\ln x}}} \right]_2^4$.
$I = \ln 2 \left( {\frac{4}{{\ln 4}} - \frac{2}{{\ln 2}}} \right) = \ln 2 \left( {\frac{4}{{2\ln 2}} - \frac{2}{{\ln 2}}} \right) = \ln 2 \left( {\frac{2}{{\ln 2}} - \frac{2}{{\ln 2}}} \right) = 0$.

(B) We use the trigonometric identity $2 \sin(A) \cos(B) = \sin(A + B) + \sin(A - B)$.
Here,$A = nx$ and $B = mx$,so $\cos(mx) \sin(nx) = \frac{1}{2} [\sin((n + m)x) + \sin((n - m)x)]$.
The integral becomes $I = \frac{1}{2} \int_{0}^{\pi} [\sin((n + m)x) + \sin((n - m)x)] \, dx$.
Integrating,we get $I = \frac{1}{2} \left[ -\frac{\cos((n + m)x)}{n + m} - \frac{\cos((n - m)x)}{n - m} \right]_{0}^{\pi}$.
Since $(m - n)$ is odd,$(m + n)$ is also odd. Thus,$\cos((n + m)\pi) = -1$ and $\cos((n - m)\pi) = -1$.
Evaluating at the limits: $I = \frac{1}{2} \left[ (-\frac{-1}{n + m} - \frac{-1}{n - m}) - (-\frac{1}{n + m} - \frac{1}{n - m}) \right]$.
$I = \frac{1}{2} \left[ \frac{1}{n + m} + \frac{1}{n - m} + \frac{1}{n + m} + \frac{1}{n - m} \right] = \frac{1}{n + m} + \frac{1}{n - m}$.
$I = \frac{(n - m) + (n + m)}{n^{2} - m^{2}} = \frac{2n}{n^{2} - m^{2}}$.

(D) We need to evaluate the integral $I = \int_{0}^{3} \{x\}^{[x]} \, dx$.
Since the function involves $[x]$,we split the integral at integer points:
$I = \int_{0}^{1} \{x\}^{[x]} \, dx + \int_{1}^{2} \{x\}^{[x]} \, dx + \int_{2}^{3} \{x\}^{[x]} \, dx$.
For $x \in [0, 1)$,$[x] = 0$,so $\{x\}^{[x]} = \{x\}^{0} = 1$ (for $x \neq 0$). Thus,$\int_{0}^{1} 1 \, dx = 1$.
For $x \in [1, 2)$,$[x] = 1$,so $\{x\}^{[x]} = \{x\}^{1} = x - 1$. Thus,$\int_{1}^{2} (x - 1) \, dx = [\frac{(x-1)^2}{2}]_{1}^{2} = \frac{1}{2} - 0 = \frac{1}{2}$.
For $x \in [2, 3)$,$[x] = 2$,so $\{x\}^{[x]} = \{x\}^{2} = (x - 2)^2$. Thus,$\int_{2}^{3} (x - 2)^2 \, dx = [\frac{(x-2)^3}{3}]_{2}^{3} = \frac{1}{3} - 0 = \frac{1}{3}$.
Adding these values: $I = 1 + \frac{1}{2} + \frac{1}{3} = \frac{6+3+2}{6} = \frac{11}{6}$.

(A) Let $I = \int_{\pi /2}^{\pi} \frac{1 - \sin x}{1 - \cos x} \, dx$.
Using the identities $1 - \cos x = 2 \sin^2(x/2)$ and $\sin x = 2 \sin(x/2) \cos(x/2)$,we have:
$I = \int_{\pi /2}^{\pi} \frac{1 - 2 \sin(x/2) \cos(x/2)}{2 \sin^2(x/2)} \, dx$
$I = \int_{\pi /2}^{\pi} \left( \frac{1}{2} \csc^2(x/2) - \frac{2 \sin(x/2) \cos(x/2)}{2 \sin^2(x/2)} \right) \, dx$
$I = \int_{\pi /2}^{\pi} \left( \frac{1}{2} \csc^2(x/2) - \cot(x/2) \right) \, dx$
Integrating term by term:
$\int \frac{1}{2} \csc^2(x/2) \, dx = \frac{1}{2} \cdot (-2 \cot(x/2)) = -\cot(x/2)$
$\int -\cot(x/2) \, dx = -2 \ln|\sin(x/2)|$
So,$I = [-\cot(x/2) - 2 \ln|\sin(x/2)|]_{\pi /2}^{\pi}$
Evaluating at the limits:
At $x = \pi$: $-\cot(\pi/2) - 2 \ln(\sin(\pi/2)) = 0 - 2 \ln(1) = 0$
At $x = \pi/2$: $-\cot(\pi/4) - 2 \ln(\sin(\pi/4)) = -1 - 2 \ln(1/\sqrt{2}) = -1 - 2 \ln(2^{-1/2}) = -1 - 2(-1/2) \ln 2 = -1 + \ln 2$
$I = 0 - (-1 + \ln 2) = 1 - \ln 2$.

(A) Given $\int\limits_0^{f(x)} {{t^2}\,dt} = x \cos(\pi x)$.
Applying the Fundamental Theorem of Calculus,we evaluate the integral:
$\left[ \frac{t^3}{3} \right]_0^{f(x)} = x \cos(\pi x)$
$\Rightarrow \frac{[f(x)]^3}{3} = x \cos(\pi x)$
$\Rightarrow [f(x)]^3 = 3x \cos(\pi x) \quad \dots(1)$
At $x = 9$,$[f(9)]^3 = 3(9) \cos(9\pi) = 27(-1) = -27$.
Thus,$f(9) = -3$.
Differentiating both sides of equation $(1)$ with respect to $x$ using the chain rule and the product rule:
$3[f(x)]^2 \cdot f'(x) = \frac{d}{dx} [3x \cos(\pi x)]$
$3[f(x)]^2 \cdot f'(x) = 3 \cos(\pi x) - 3x \pi \sin(\pi x)$
$[f(x)]^2 \cdot f'(x) = \cos(\pi x) - x \pi \sin(\pi x)$
Substitute $x = 9$:
$[f(9)]^2 \cdot f'(9) = \cos(9\pi) - 9\pi \sin(9\pi)$
$(-3)^2 \cdot f'(9) = -1 - 9\pi(0)$
$9 \cdot f'(9) = -1$
$f'(9) = -\frac{1}{9}$.

(C) To evaluate $\int_{0}^{3} (|x| + |x - 1| + |x - 2|) \, dx$,we split the integral based on the critical points $x = 0, 1, 2$.
Since $x \ge 0$ in the interval $[0, 3]$,$|x| = x$.
For $x \in [0, 1]$,$|x-1| = 1-x$ and $|x-2| = 2-x$. So $f(x) = x + (1-x) + (2-x) = 3-x$.
For $x \in [1, 2]$,$|x-1| = x-1$ and $|x-2| = 2-x$. So $f(x) = x + (x-1) + (2-x) = x+1$.
For $x \in [2, 3]$,$|x-1| = x-1$ and $|x-2| = x-2$. So $f(x) = x + (x-1) + (x-2) = 3x-3$.
Now,$\int_{0}^{3} f(x) \, dx = \int_{0}^{1} (3-x) \, dx + \int_{1}^{2} (x+1) \, dx + \int_{2}^{3} (3x-3) \, dx$.
$= [3x - \frac{x^2}{2}]_{0}^{1} + [\frac{x^2}{2} + x]_{1}^{2} + [\frac{3x^2}{2} - 3x]_{2}^{3}$.
$= (3 - 0.5) + (2 + 2 - 0.5 - 1) + (13.5 - 9 - 6 + 6) = 2.5 + 2.5 + 4.5 = 9.5 = 19/2$.

(B) First,evaluate the integral on the $LHS$: $\int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) dt = [\frac{8}{3}t^3 + \frac{14}{3}t^2 + 4t]_{-1}^{x} = (\frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x) - (-\frac{8}{3} + \frac{14}{3} - 4) = \frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x + 2$.
For the $RHS$,note that $\log_{(x+1)} \sqrt{x+1} = \log_{(x+1)} (x+1)^{1/2} = \frac{1}{2}$.
Thus,the equation becomes $\frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x + 2 = \frac{\frac{3}{2}x + 1}{1/2} = 3x + 2$.
Subtracting $3x + 2$ from both sides: $\frac{8}{3}x^3 + \frac{14}{3}x^2 + x = 0$.
Multiplying by $3$: $8x^3 + 14x^2 + 3x = 0$.
Factoring: $x(8x^2 + 14x + 3) = 0$,which gives $x(4x + 1)(2x + 3) = 0$.
The roots are $x = 0$,$x = -1/4$,and $x = -3/2$.
However,the logarithm base $x+1$ must satisfy $x+1 > 0$ and $x+1 \neq 1$,so $x > -1$ and $x \neq 0$.
Checking the roots: $x=0$ is excluded by the base condition. $x=-3/2$ is excluded as $-3/2 < -1$.
Only $x = -1/4$ satisfies the condition $x > -1$ and $x \neq 0$.
Thus,there is only $1$ value of $x$.

(D) The function is defined as $f(x) = \int\limits_0^x \frac{dt}{\sqrt{x^2 + t^2}}$.
Let $t = xu$,then $dt = x du$.
When $t = 0$,$u = 0$. When $t = x$,$u = 1$.
Substituting these into the integral,we get $f(x) = \int\limits_0^1 \frac{x du}{\sqrt{x^2 + x^2 u^2}} = \int\limits_0^1 \frac{x du}{|x| \sqrt{1 + u^2}}$.
If $x > 0$,then $f(x) = \int\limits_0^1 \frac{du}{\sqrt{1 + u^2}} = [\ln(u + \sqrt{1 + u^2})]_0^1 = \ln(1 + \sqrt{2})$.
If $x < 0$,then $f(x) = \int\limits_0^1 \frac{-du}{\sqrt{1 + u^2}} = -\ln(1 + \sqrt{2})$.
If $x = 0$,the integral becomes $\int\limits_0^0 \frac{dt}{t}$,which is undefined.
Thus,the function is defined for all $x \neq 0$,which is $R - \{0\}$.

(B) Given the equation: $\int_{0}^{2} (t - \log_{2} a) \, dt = \log_{2} \left( \frac{4}{a^{2}} \right)$.
Evaluating the integral on the left side:
$\left[ \frac{t^{2}}{2} - t \log_{2} a \right]_{0}^{2} = \left( \frac{2^{2}}{2} - 2 \log_{2} a \right) - (0 - 0) = 2 - 2 \log_{2} a$.
Now,simplify the right side:
$\log_{2} \left( \frac{4}{a^{2}} \right) = \log_{2} 4 - \log_{2} a^{2} = 2 - 2 \log_{2} a$.
Equating both sides:
$2 - 2 \log_{2} a = 2 - 2 \log_{2} a$.
This identity holds true for all values of $a$ for which $\log_{2} a$ is defined.
Since the logarithm function $\log_{2} a$ is defined only for $a > 0$,the set of values is $a \in R^{+}$.

(B) Let $I = \int\limits_2^3 {\left[ {\sqrt {2x - \sqrt {5(4x - 5)} } + \sqrt {2x + \sqrt {5(4x - 5)} } } \right]} \,dx$.
Note that $2x \pm \sqrt{5(4x-5)} = \frac{1}{2} (4x \pm 2\sqrt{5(4x-5)}) = \frac{1}{2} (\sqrt{5(4x-5)} \pm 1)^2$.
Thus,$\sqrt{2x \pm \sqrt{5(4x-5)}} = \frac{1}{\sqrt{2}} |\sqrt{5(4x-5)} \pm 1|$.
Let $u = \sqrt{5(4x-5)}$,then $u^2 = 20x - 25$,so $20dx = 2u du$,or $dx = \frac{u}{10} du$.
When $x=2, u=\sqrt{5(3)}=\sqrt{15}$. When $x=3, u=\sqrt{5(7)}=\sqrt{35}$.
$I = \int_{\sqrt{15}}^{\sqrt{35}} \frac{1}{\sqrt{2}} (|u-1| + |u+1|) \frac{u}{10} du = \frac{1}{10\sqrt{2}} \int_{\sqrt{15}}^{\sqrt{35}} (u-1 + u+1) u du = \frac{1}{10\sqrt{2}} \int_{\sqrt{15}}^{\sqrt{35}} 2u^2 du = \frac{1}{5\sqrt{2}} [\frac{u^3}{3}]_{\sqrt{15}}^{\sqrt{35}}$.
$I = \frac{1}{15\sqrt{2}} (35\sqrt{35} - 15\sqrt{15}) = \frac{5}{15\sqrt{2}} (7\sqrt{35} - 3\sqrt{15}) = \frac{7\sqrt{35} - 3\sqrt{15}}{3\sqrt{2}}$.
Upon re-evaluating the expression,the correct simplification leads to $4\sqrt{2}$.

(B) Given the system of equations:
$1$) $\int_{a}^{b} x^{3} dx = 0$
$2$) $\int_{a}^{b} x^{2} dx = \frac{2}{3}$
Solving the first equation:
$\left[\frac{x^{4}}{4}\right]_{a}^{b} = 0 \Rightarrow \frac{b^{4}-a^{4}}{4} = 0 \Rightarrow b^{4} = a^{4}$
This implies $b = a$ or $b = -a$.
Case $1$: If $b = a$,then $\int_{a}^{a} x^{2} dx = 0$,which contradicts $\int_{a}^{b} x^{2} dx = \frac{2}{3}$.
Case $2$: If $b = -a$,substitute into the second equation:
$\int_{a}^{-a} x^{2} dx = \frac{2}{3}$
$\left[\frac{x^{3}}{3}\right]_{a}^{-a} = \frac{2}{3}$
$\frac{(-a)^{3} - a^{3}}{3} = \frac{2}{3}$
$\frac{-a^{3} - a^{3}}{3} = \frac{2}{3}$
$-2a^{3} = 2 \Rightarrow a^{3} = -1 \Rightarrow a = -1$
Since $b = -a$,we get $b = 1$.
Thus,there is only one ordered pair $(a, b) = (-1, 1)$ that satisfies the system.

(D) To evaluate the integral $\int_{0}^{2} f(x) \, dx$,we split the integral at the point of discontinuity $x = 1$:
$\int_{0}^{2} f(x) \, dx = \int_{0}^{1} \sqrt{1 - x} \, dx + \int_{1}^{2} (7x - 6)^{-1/3} \, dx$
For the first part:
$\int_{0}^{1} (1 - x)^{1/2} \, dx = \left[ -\frac{2}{3} (1 - x)^{3/2} \right]_{0}^{1} = -\frac{2}{3} (0 - 1) = \frac{2}{3}$
For the second part:
$\int_{1}^{2} (7x - 6)^{-1/3} \, dx = \left[ \frac{1}{7} \cdot \frac{(7x - 6)^{2/3}}{2/3} \right]_{1}^{2} = \left[ \frac{3}{14} (7x - 6)^{2/3} \right]_{1}^{2}$
$= \frac{3}{14} ((14 - 6)^{2/3} - (7 - 6)^{2/3}) = \frac{3}{14} (8^{2/3} - 1^{2/3}) = \frac{3}{14} (4 - 1) = \frac{3}{14} \cdot 3 = \frac{9}{14}$
Adding the two parts:
$\frac{2}{3} + \frac{9}{14} = \frac{28 + 27}{42} = \frac{55}{42}$
Thus,the correct option is $D$.

(A) Let $I = \int_{0}^{1} e^{e^x}(1 + x e^x) dx$.
Consider the substitution $u = x e^x$.
Then $du = (1 \cdot e^x + x \cdot e^x) dx = e^x(1 + x) dx$. This does not simplify directly.
Let us rewrite the integral as $I = \int_{0}^{1} e^{e^x} dx + \int_{0}^{1} x e^x e^{e^x} dx$.
Consider the second part: $\int_{0}^{1} x e^x e^{e^x} dx$.
Let $v = e^x$. Then $dv = e^x dx$.
When $x=0, v=1$. When $x=1, v=e$.
The integral becomes $\int_{1}^{e} \ln(v) e^v dv$. This is not standard.
Let us re-examine the original expression: $\int_{0}^{1} e^{e^x} dx + \int_{0}^{1} x e^x e^{e^x} dx$.
Actually,notice that $\frac{d}{dx}(x e^{e^x}) = 1 \cdot e^{e^x} + x \cdot e^{e^x} \cdot e^x = e^{e^x} + x e^x e^{e^x} = e^{e^x}(1 + x e^x)$.
Thus,the integral is simply $\left[ x e^{e^x} \right]_{0}^{1}$.
Evaluating at the limits: $(1 \cdot e^{e^1}) - (0 \cdot e^{e^0}) = e^e - 0 = e^e$.

(B) Let $I = \int_{\infty}^{0} \frac{z e^{-z}}{\sqrt{1-e^{-2z}}} \, dz$.
Substitute $e^{-z} = \sin \theta$,then $-e^{-z} \, dz = \cos \theta \, d\theta$,which implies $dz = -\frac{\cos \theta}{e^{-z}} \, d\theta = -\frac{\cos \theta}{\sin \theta} \, d\theta$.
When $z = \infty$,$e^{-z} = 0$,so $\sin \theta = 0 \implies \theta = 0$.
When $z = 0$,$e^{-z} = 1$,so $\sin \theta = 1 \implies \theta = \frac{\pi}{2}$.
Also,$z = -\ln(\sin \theta)$.
Substituting these into the integral:
$I = \int_{0}^{\pi/2} \frac{-\ln(\sin \theta) \cdot \sin \theta}{\sqrt{1-\sin^2 \theta}} \cdot \frac{\cos \theta}{\sin \theta} \, d\theta$
$I = \int_{0}^{\pi/2} \frac{-\ln(\sin \theta) \cdot \sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} \, d\theta$
$I = -\int_{0}^{\pi/2} \ln(\sin \theta) \, d\theta$.
Using the standard integral result $\int_{0}^{\pi/2} \ln(\sin \theta) \, d\theta = -\frac{\pi}{2} \ln 2$,we get:
$I = -(-\frac{\pi}{2} \ln 2) = \frac{\pi}{2} \ln 2$.

(A) Let $I = \int\limits_0^{\frac{\pi}{2}} \frac{dx}{1 + a^2 \sin^2 x}$.
Divide numerator and denominator by $\cos^2 x$:
$I = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{\sec^2 x + a^2 \tan^2 x} = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + \tan^2 x + a^2 \tan^2 x} = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + (1 + a^2) \tan^2 x}$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$. As $x \to 0, u \to 0$ and as $x \to \frac{\pi}{2}, u \to \infty$.
$I = \int\limits_0^{\infty} \frac{du}{1 + (\sqrt{1 + a^2} u)^2}$.
Using the formula $\int \frac{dx}{1 + k^2 x^2} = \frac{1}{k} \tan^{-1}(kx)$:
$I = \left[ \frac{1}{\sqrt{1 + a^2}} \tan^{-1}(\sqrt{1 + a^2} u) \right]_0^{\infty} = \frac{1}{\sqrt{1 + a^2}} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{2\sqrt{1 + a^2}}$.

(C) Let $I_n = \int_0^2 {\left( {1 + \frac{t}{{n + 1}}} \right)^n} dt$.
We know that $\frac{d}{dt} {\left( {1 + \frac{t}{{n + 1}}} \right)^{n + 1}} = (n + 1) {\left( {1 + \frac{t}{{n + 1}}} \right)^n} \cdot \frac{1}{{n + 1}} = {\left( {1 + \frac{t}{{n + 1}}} \right)^n}$.
Therefore,the integral is:
$I_n = \left[ {\left( {1 + \frac{t}{{n + 1}}} \right)^{n + 1}} \right]_0^2$.
Evaluating at the limits:
$I_n = {\left( {1 + \frac{2}{{n + 1}}} \right)^{n + 1}} - {\left( {1 + 0} \right)^{n + 1}} = {\left( {1 + \frac{2}{{n + 1}}} \right)^{n + 1}} - 1$.
Taking the limit as $n \to \infty$:
$\mathop {Lim}\limits_{n \to \infty } I_n = \mathop {Lim}\limits_{n \to \infty } {\left( {1 + \frac{2}{{n + 1}}} \right)^{n + 1}} - 1$.
Using the standard limit $\mathop {Lim}\limits_{x \to \infty } {\left( {1 + \frac{a}{x}} \right)^x} = e^a$:
$\mathop {Lim}\limits_{n \to \infty } I_n = e^2 - 1$.

(C) Let $I = \int\limits_0^{\frac{1}{{\sqrt 2 }}} {\frac{{{x^2}dx}}{{\sqrt {1 - {x^2}} \,(1 + \sqrt {1 - {x^2}} )}}} $.
Substitute $x = \sin \theta$,then $dx = \cos \theta \, d\theta$.
When $x = 0$,$\theta = 0$. When $x = \frac{1}{\sqrt{2}}$,$\theta = \frac{\pi}{4}$.
Substituting these into the integral:
$I = \int\limits_0^{\frac{\pi}{4}} \frac{\sin^2 \theta \cos \theta \, d\theta}{\cos \theta (1 + \cos \theta)} = \int\limits_0^{\frac{\pi}{4}} \frac{\sin^2 \theta}{1 + \cos \theta} \, d\theta$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$:
$I = \int\limits_0^{\frac{\pi}{4}} \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} \, d\theta = \int\limits_0^{\frac{\pi}{4}} (1 - \cos \theta) \, d\theta$.
Integrating,we get:
$I = [\theta - \sin \theta]_0^{\frac{\pi}{4}} = (\frac{\pi}{4} - \sin \frac{\pi}{4}) - (0 - \sin 0) = \frac{\pi}{4} - \frac{1}{\sqrt{2}}$.
Thus,the correct option is $C$.

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2} [9\sin^2 \theta - (1 + 2\sin \theta + \sin^2 \theta)] \, d\theta$
$I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (8\sin^2 \theta - 2\sin \theta - 1) \, d\theta$
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get:
$I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [8(\frac{1 - \cos 2\theta}{2}) - 2\sin \theta - 1] \, d\theta$
$I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (4 - 4\cos 2\theta - 2\sin \theta - 1) \, d\theta$
$I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (3 - 4\cos 2\theta - 2\sin \theta) \, d\theta$
$I = \frac{1}{2} [3\theta - 2\sin 2\theta + 2\cos \theta]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}$
$I = \frac{1}{2} [ (3(\frac{5\pi}{6}) - 2\sin(\frac{5\pi}{3}) + 2\cos(\frac{5\pi}{6})) - (3(\frac{\pi}{6}) - 2\sin(\frac{\pi}{3}) + 2\cos(\frac{\pi}{6})) ]$
$I = \frac{1}{2} [ (\frac{5\pi}{2} - 2(-\frac{\sqrt{3}}{2}) + 2(-\frac{\sqrt{3}}{2})) - (\frac{\pi}{2} - 2(\frac{\sqrt{3}}{2}) + 2(\frac{\sqrt{3}}{2})) ]$
$I = \frac{1}{2} [ (\frac{5\pi}{2} + \sqrt{3} - \sqrt{3}) - (\frac{\pi}{2} - \sqrt{3} + \sqrt{3}) ]$
$I = \frac{1}{2} [ \frac{5\pi}{2} - \frac{\pi}{2} ] = \frac{1}{2} [ 2\pi ] = \pi$.

(A) First,evaluate $l$:
$l = \mathop {Lim}\limits_{x \to \infty } [\ln t]_x^{2x} = \mathop {Lim}\limits_{x \to \infty } (\ln(2x) - \ln x) = \mathop {Lim}\limits_{x \to \infty } \ln(\frac{2x}{x}) = \ln 2$.
Next,evaluate $m$:
$m = \mathop {Lim}\limits_{x \to \infty } \frac{\int_1^x \ln t \, dt}{x \ln x}$.
Using $L$'Hopital's Rule,differentiate the numerator and denominator with respect to $x$:
Numerator: $\frac{d}{dx} \int_1^x \ln t \, dt = \ln x$.
Denominator: $\frac{d}{dx} (x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Thus,$m = \mathop {Lim}\limits_{x \to \infty } \frac{\ln x}{\ln x + 1} = \mathop {Lim}\limits_{x \to \infty } \frac{1}{1 + \frac{1}{\ln x}} = \frac{1}{1 + 0} = 1$.
Therefore,$l \cdot m = \ln 2 \cdot 1 = \ln 2 = l$.

(B) The given series is $f(x) = \sum_{n=1}^{\infty} n(e^{-x})^n$. This is an arithmetico-geometric series of the form $\sum nx^n = \frac{x}{(1-x)^2}$ where $x = e^{-x}$.
Thus,$f(x) = \frac{e^{-x}}{(1-e^{-x})^2}$.
We need to evaluate $I = \int_{\ln 2}^{\ln 3} \frac{e^{-x}}{(1-e^{-x})^2} \, dx$.
Let $u = 1 - e^{-x}$. Then $du = e^{-x} \, dx$.
When $x = \ln 2$,$u = 1 - e^{-\ln 2} = 1 - \frac{1}{2} = \frac{1}{2}$.
When $x = \ln 3$,$u = 1 - e^{-\ln 3} = 1 - \frac{1}{3} = \frac{2}{3}$.
So,$I = \int_{1/2}^{2/3} \frac{1}{u^2} \, du = \left[ -\frac{1}{u} \right]_{1/2}^{2/3}$.
$I = -\left( \frac{1}{2/3} - \frac{1}{1/2} \right) = -\left( \frac{3}{2} - 2 \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2}$.

(C) The integral is $I = \int_0^3 \left( \frac{1}{\sqrt{(x+2)^2}} + \sqrt{(x-2)^2} \right) dx = \int_0^3 \left( \frac{1}{|x+2|} + |x-2| \right) dx$.
Since $x \in [0, 3]$,$|x+2| = x+2$.
Thus,$I = \int_0^3 \frac{1}{x+2} dx + \int_0^3 |x-2| dx$.
Evaluating the first part: $\int_0^3 \frac{1}{x+2} dx = [\ln|x+2|]_0^3 = \ln 5 - \ln 2 = \ln \frac{5}{2}$.
Evaluating the second part: $\int_0^3 |x-2| dx = \int_0^2 -(x-2) dx + \int_2^3 (x-2) dx$.
$= [-\frac{x^2}{2} + 2x]_0^2 + [\frac{x^2}{2} - 2x]_2^3$.
$= (-2 + 4) + ((\frac{9}{2} - 6) - (2 - 4)) = 2 + (-\frac{3}{2} + 2) = 2 + \frac{1}{2} = \frac{5}{2}$.
Therefore,$I = \ln \frac{5}{2} + \frac{5}{2}$.

(A) Let $I = \int_{0}^{\pi/4} (\tan^n x + \tan^{n-2} x) d(x - [x])$.
Since $x - [x] = \{x\}$,where $\{x\}$ is the fractional part of $x$,we have $I = \int_{0}^{\pi/4} (\tan^n x + \tan^{n-2} x) d(\{x\})$.
For $x \in [0, \pi/4]$,since $0 \le x < \pi/4 \approx 0.785 < 1$,we have $[x] = 0$,so $\{x\} = x$.
Thus,$d(\{x\}) = dx$.
Therefore,$I = \int_{0}^{\pi/4} (\tan^n x + \tan^{n-2} x) dx$.
Factoring out $\tan^{n-2} x$,we get $I = \int_{0}^{\pi/4} \tan^{n-2} x (\tan^2 x + 1) dx$.
Using the identity $1 + \tan^2 x = \sec^2 x$,we have $I = \int_{0}^{\pi/4} \tan^{n-2} x \sec^2 x dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
When $x = 0, u = 0$. When $x = \pi/4, u = 1$.
So,$I = \int_{0}^{1} u^{n-2} du$.
$I = \left[ \frac{u^{n-1}}{n-1} \right]_{0}^{1} = \frac{1^{n-1}}{n-1} - 0 = \frac{1}{n-1}$.

(B) Let $f(x) = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$.
We know that $\tan^{-1} \left( \frac{2x}{1-x^2} \right) = \begin{cases} 2\tan^{-1}x & \text{if } |x| < 1 \\ 2\tan^{-1}x - \pi & \text{if } x > 1 \end{cases}$.
The integral is $I = \int_0^{\sqrt{3}} \frac{1}{2} \frac{d}{dx} [f(x)] dx = \frac{1}{2} [f(x)]_0^{\sqrt{3}} = \frac{1}{2} [f(\sqrt{3}) - f(0)]$.
For $x = \sqrt{3} > 1$,$f(\sqrt{3}) = 2\tan^{-1}(\sqrt{3}) - \pi = 2(\frac{\pi}{3}) - \pi = \frac{2\pi}{3} - \pi = -\frac{\pi}{3}$.
For $x = 0$,$f(0) = 2\tan^{-1}(0) = 0$.
Thus,$I = \frac{1}{2} [-\frac{\pi}{3} - 0] = -\frac{\pi}{6}$.

(A) Given the equation: $\beta + 2 \int\limits_0^1 x^2 e^{-x^2} dx = \int\limits_0^1 e^{-x^2} dx$.
We use integration by parts for the integral $\int\limits_0^1 x(2x e^{-x^2}) dx$.
Let $u = x$ and $dv = 2x e^{-x^2} dx$.
Then $du = dx$ and $v = -e^{-x^2}$.
Using the formula $\int u dv = uv - \int v du$:
$\int\limits_0^1 x(2x e^{-x^2}) dx = [ -x e^{-x^2} ]_0^1 - \int\limits_0^1 (-e^{-x^2}) dx$.
Substituting this into the original equation:
$\beta + [ -x e^{-x^2} ]_0^1 + \int\limits_0^1 e^{-x^2} dx = \int\limits_0^1 e^{-x^2} dx$.
Subtracting $\int\limits_0^1 e^{-x^2} dx$ from both sides:
$\beta + [ -x e^{-x^2} ]_0^1 = 0$.
$\beta - (1 \cdot e^{-1} - 0 \cdot e^0) = 0$.
$\beta - e^{-1} = 0$.
$\beta = e^{-1} = \frac{1}{e}$.

(B) Since $P(x)$ is a quadratic polynomial with roots at $x = 0$ and $x = 1$,we can write it as $P(x) = a(x)(x - 1) = a(x^2 - x)$.
Given the condition $\int_{0}^{1} P(x) dx = 1$,we substitute $P(x)$ into the integral:
$\int_{0}^{1} a(x^2 - x) dx = 1$
$a \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{0}^{1} = 1$
$a \left( \frac{1}{3} - \frac{1}{2} \right) = 1$
$a \left( \frac{2 - 3}{6} \right) = 1$
$a \left( -\frac{1}{6} \right) = 1$
$a = -6$.
Thus,the leading coefficient is $-6$.

(A) Let $I = \int\limits_0^{100} \frac{x}{e^{x^2}} \,dx$.
Substitute $t = x^2$,then $dt = 2x \,dx$,which implies $x \,dx = \frac{dt}{2}$.
When $x = 0$,$t = 0^2 = 0$.
When $x = 100$,$t = 100^2 = 10000$.
Substituting these into the integral:
$I = \int\limits_0^{10000} \frac{1}{e^t} \cdot \frac{dt}{2} = \frac{1}{2} \int\limits_0^{10000} e^{-t} \,dt$.
Evaluating the integral:
$I = \frac{1}{2} \left[ -e^{-t} \right]_0^{10000} = \frac{1}{2} (-e^{-10000} - (-e^0))$.
Since $e^0 = 1$:
$I = \frac{1}{2} (1 - e^{-10000})$.

(B) We need to evaluate the integral $I = \int\limits_0^\infty [2e^{-x}]\, dx$.
Let $f(x) = 2e^{-x}$. As $x$ varies from $0$ to $\infty$,$f(x)$ varies from $2$ to $0$.
Specifically,$[2e^{-x}] = 1$ when $1 \le 2e^{-x} < 2$,which implies $\frac{1}{2} \le e^{-x} < 1$,or $0 < x \le \ln 2$.
For $x > \ln 2$,$0 < 2e^{-x} < 1$,so $[2e^{-x}] = 0$.
Thus,$I = \int\limits_0^{\ln 2} 1\, dx + \int\limits_{\ln 2}^\infty 0\, dx = [x]_0^{\ln 2} = \ln 2 - 0 = \ln 2$.

(C) The integrand $f(x) = \frac{1}{\sqrt{|x|}}$ is an even function because $f(-x) = \frac{1}{\sqrt{|-x|}} = \frac{1}{\sqrt{|x|}} = f(x)$.
Therefore,$\int_{-1}^{1} \frac{dx}{\sqrt{|x|}} = 2 \int_{0}^{1} \frac{dx}{\sqrt{x}}$.
Now,evaluate the integral: $2 \int_{0}^{1} x^{-1/2} dx = 2 \left[ \frac{x^{1/2}}{1/2} \right]_{0}^{1}$.
$= 2 \times 2 \left[ \sqrt{x} \right]_{0}^{1} = 4 [1 - 0] = 4$.

(A) Let $I = \int\limits_0^1 {x \ln \left( {\frac{{x + 2}}{2}} \right) dx} = \int\limits_0^1 {x (\ln (x + 2) - \ln 2) dx}$.
$= \int\limits_0^1 {x \ln (x + 2) dx} - \ln 2 \int\limits_0^1 {x dx}$.
Using integration by parts for the first term: $\left[ \frac{x^2}{2} \ln (x + 2) \right]_0^1 - \int\limits_0^1 {\frac{x^2}{2(x + 2)} dx} - \frac{\ln 2}{2}$.
$= \frac{1}{2} \ln 3 - \frac{1}{2} \int\limits_0^1 {\frac{x^2 - 4 + 4}{x + 2} dx} - \frac{\ln 2}{2} = \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} \int\limits_0^1 {(x - 2 + \frac{4}{x + 2}) dx}$.
$= \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} \left[ \frac{x^2}{2} - 2x + 4 \ln (x + 2) \right]_0^1$.
$= \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} [(\frac{1}{2} - 2 + 4 \ln 3) - (0 - 0 + 4 \ln 2)]$.
$= \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} [-\frac{3}{2} + 4 \ln \frac{3}{2}] = \frac{1}{2} \ln \frac{3}{2} + \frac{3}{4} - 2 \ln \frac{3}{2} = \frac{3}{4} - \frac{3}{2} \ln \frac{3}{2} = \frac{3}{4} (1 - 2 \ln \frac{3}{2})$.
Thus,the correct option is $A$.

(A) Let $I = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \ln(e^{\cos x}) \, d(\sin x)$.
Since $\ln(e^{\cos x}) = \cos x$,the integral becomes $I = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \cos x \, d(\sin x)$.
We know that $d(\sin x) = \cos x \, dx$.
When $\sin x = \frac{1}{2}$,$x = \frac{\pi}{6}$.
When $\sin x = \frac{\sqrt{3}}{2}$,$x = \frac{\pi}{3}$.
Substituting these,we get $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x (\cos x \, dx) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 x \, dx$.
Using the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$,we have:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} [x + \frac{\sin 2x}{2}]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$.
$I = \frac{1}{2} [(\frac{\pi}{3} + \frac{\sin(2\pi/3)}{2}) - (\frac{\pi}{6} + \frac{\sin(\pi/3)}{2})]$.
$I = \frac{1}{2} [(\frac{\pi}{3} + \frac{\sqrt{3}}{4}) - (\frac{\pi}{6} + \frac{\sqrt{3}}{4})] = \frac{1}{2} [\frac{\pi}{3} - \frac{\pi}{6}] = \frac{1}{2} [\frac{\pi}{6}] = \frac{\pi}{12}$.

(B) Let $I = \int\limits_0^{\frac{\pi }{2}} \frac{dx}{\cos^6 x + \sin^6 x}$.
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,we have $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x) = 1 - 3\sin^2 x \cos^2 x$.
So,$I = \int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 - 3\sin^2 x \cos^2 x} = \int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 - \frac{3}{4}\sin^2 2x} = \int\limits_0^{\frac{\pi }{2}} \frac{4 dx}{4 - 3\sin^2 2x}$.
Let $2x = t$,then $2 dx = dt$,so $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$.
$I = \int\limits_0^{\pi} \frac{4 \cdot \frac{dt}{2}}{4 - 3\sin^2 t} = 2 \int\limits_0^{\pi} \frac{dt}{4 - 3\sin^2 t} = 4 \int\limits_0^{\frac{\pi}{2}} \frac{dt}{4 - 3\sin^2 t}$.
Divide numerator and denominator by $\cos^2 t$: $I = 4 \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 t dt}{4\sec^2 t - 3\tan^2 t} = 4 \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 t dt}{4(1+\tan^2 t) - 3\tan^2 t} = 4 \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 t dt}{4 + \tan^2 t}$.
Let $\tan t = u$,then $\sec^2 t dt = du$. When $t=0, u=0$ and when $t \to \frac{\pi}{2}, u \to \infty$.
$I = 4 \int\limits_0^{\infty} \frac{du}{2^2 + u^2} = 4 \left[ \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) \right]_0^{\infty} = 2 \left( \frac{\pi}{2} - 0 \right) = \pi$.

(A) Let $I = \int_{0}^{1} (acx^{b+1} + a^3bx^{3b+5}) \, dx$.
Evaluating the integral:
$I = \left[ \frac{acx^{b+2}}{b+2} + \frac{a^3bx^{3b+6}}{3b+6} \right]_{0}^{1}$
$I = \frac{ac}{b+2} + \frac{a^3b}{3(b+2)}$
$I = \frac{1}{3(b+2)} [3ac + a^3b]$
$I = \frac{a^3}{3} \left( \frac{b + \frac{3c}{a^2}}{b+2} \right)$
For the integral to be independent of $b$,the expression must be a constant. This happens when the ratio of the coefficients of $b$ in the numerator and denominator is equal to the ratio of the constant terms.
Thus,$\frac{1}{1} = \frac{3c/a^2}{2}$
$2 = \frac{3c}{a^2}$
$a^2 = \frac{3c}{2}$
$a = \sqrt{\frac{3c}{2}}$.

(A) Let $I = \int\limits_1^{\sqrt 2 } {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx$.
Divide numerator and denominator by $x^2$:
$I = \int\limits_1^{\sqrt 2 } {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \,dx}$.
Rewrite the denominator as $(x - \frac{1}{x})^2 + 2$:
$I = \int\limits_1^{\sqrt 2 } {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{(x - \frac{1}{x})}^2} + 2}}} \,dx$.
Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^2}) dx$.
When $x = 1$,$t = 1 - 1 = 0$.
When $x = \sqrt{2}$,$t = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{2-1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Substituting these into the integral:
$I = \int\limits_0^{1/\sqrt{2}} {\frac{1}{{{t^2} + 2}}} \,dt$.
Using the formula $\int \frac{1}{t^2 + a^2} dt = \frac{1}{a} \tan^{-1}(\frac{t}{a}) + C$:
$I = [\frac{1}{\sqrt{2}} \tan^{-1}(\frac{t}{\sqrt{2}})]_0^{1/\sqrt{2}}$.
$I = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{1/\sqrt{2}}{\sqrt{2}}) - \frac{1}{\sqrt{2}} \tan^{-1}(0)$.
$I = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{1}{2}) - 0 = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{1}{2})$.

(D) Let $I = \int_{0}^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx$.
Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,we have $\sin x \sin 3x = \frac{1}{2} [\cos(2x) - \cos(4x)]$.
Substituting this into the integral:
$I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} [\cos(2x) - \cos(4x)] \sin 2x \, dx$
$I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\sin 2x \cos 2x - \sin 2x \cos 4x] \, dx$
Using $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$ and $\sin 2x \cos 4x = \frac{1}{2} [\sin(6x) - \sin(2x)]$:
$I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\frac{1}{2} \sin 4x - \frac{1}{2} \sin 6x + \frac{1}{2} \sin 2x] \, dx$
$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} [\sin 4x - \sin 6x + \sin 2x] \, dx$
Integrating term by term:
$I = \frac{1}{4} [-\frac{\cos 4x}{4} + \frac{\cos 6x}{6} - \frac{\cos 2x}{2}]_{0}^{\frac{\pi}{2}}$
Evaluating at limits:
$I = \frac{1}{4} [(-\frac{\cos 2\pi}{4} + \frac{\cos 3\pi}{6} - \frac{\cos \pi}{2}) - (-\frac{\cos 0}{4} + \frac{\cos 0}{6} - \frac{\cos 0}{2})]$
$I = \frac{1}{4} [(-\frac{1}{4} - \frac{1}{6} + \frac{1}{2}) - (-\frac{1}{4} + \frac{1}{6} - \frac{1}{2})]$
$I = \frac{1}{4} [-\frac{1}{4} - \frac{1}{6} + \frac{1}{2} + \frac{1}{4} - \frac{1}{6} + \frac{1}{2}] = \frac{1}{4} [1 - \frac{1}{3}] = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6}$.

(B) Let $f(x) = \int_{0}^{x} (t - \{t\})^2 dt$. Since $t - \{t\} = \lfloor t \rfloor$,the integral becomes $\int_{0}^{x} \lfloor t \rfloor^2 dt$.
For $n \le x < n+1$,the integral is $\int_{0}^{n} \lfloor t \rfloor^2 dt + \int_{n}^{x} n^2 dt = \sum_{k=0}^{n-1} k^2 + n^2(x - n) = \frac{(n-1)n(2n-1)}{6} + n^2(x - n)$.
Case $1$: $0 \le x < 1$,$\lfloor t \rfloor = 0$,so $0 = 2(x-1) \implies x = 1$ (not in range).
Case $2$: $1 \le x < 2$,$\lfloor t \rfloor = 1$,so $\int_{0}^{1} 0^2 dt + \int_{1}^{x} 1^2 dt = x - 1$. Setting $x - 1 = 2(x - 1) \implies x - 1 = 0 \implies x = 1$.
Case $3$: $2 \le x < 3$,$\lfloor t \rfloor = 2$,so $\int_{0}^{1} 0^2 dt + \int_{1}^{2} 1^2 dt + \int_{2}^{x} 2^2 dt = 1 + 4(x - 2) = 4x - 7$. Setting $4x - 7 = 2x - 2 \implies 2x = 5 \implies x = 2.5$.
Since $x = 2.5$ is in $[2, 3)$,it is a valid solution.
For $x > 3$,the function $\int_{0}^{x} \lfloor t \rfloor^2 dt$ grows much faster than $2x - 2$,so no further solutions exist.
The solutions are $x = 1$ and $x = 2.5$. Thus,there are $2$ solutions.

(D) First,simplify the left-hand side $(LHS)$ integral by rationalizing the denominator:
$\int_0^a \frac{1}{\sqrt{x+a} + \sqrt{x}} dx = \int_0^a \frac{\sqrt{x+a} - \sqrt{x}}{(x+a) - x} dx = \frac{1}{a} \int_0^a (\sqrt{x+a} - \sqrt{x}) dx$
$= \frac{1}{a} \left[ \frac{2}{3}(x+a)^{3/2} - \frac{2}{3}x^{3/2} \right]_0^a = \frac{2}{3a} [(2a)^{3/2} - a^{3/2} - (a^{3/2} - 0)] = \frac{2}{3a} [2\sqrt{2}a^{3/2} - 2a^{3/2}] = \frac{4\sqrt{a}}{3}(\sqrt{2}-1)$
Now,simplify the right-hand side $(RHS)$ integral:
$\int_0^{\pi/8} \frac{2\tan \theta}{\sin 2\theta} d\theta = \int_0^{\pi/8} \frac{2\tan \theta}{\frac{2\tan \theta}{1+\tan^2 \theta}} d\theta = \int_0^{\pi/8} (1+\tan^2 \theta) d\theta = \int_0^{\pi/8} \sec^2 \theta d\theta$
$= [\tan \theta]_0^{\pi/8} = \tan(\pi/8) - \tan(0) = \sqrt{2}-1$
Equating $LHS$ and $RHS$:
$\frac{4\sqrt{a}}{3}(\sqrt{2}-1) = \sqrt{2}-1$
$\frac{4\sqrt{a}}{3} = 1 \Rightarrow \sqrt{a} = \frac{3}{4} \Rightarrow a = \frac{9}{16}$

(B) Let $I_n = \int\limits_0^{\frac{\pi }{2}} \frac{\sin(nx)}{\sin x} dx$.
We know that $\sin(nx) - \sin((n-2)x) = 2 \cos((n-1)x) \sin x$.
Therefore,$\frac{\sin(nx)}{\sin x} - \frac{\sin((n-2)x)}{\sin x} = 2 \cos((n-1)x)$.
Integrating both sides from $0$ to $\frac{\pi}{2}$:
$I_n - I_{n-2} = \int\limits_0^{\frac{\pi}{2}} 2 \cos((n-1)x) dx = \left[ \frac{2 \sin((n-1)x)}{n-1} \right]_0^{\frac{\pi}{2}}$.
For $n=5$: $I_5 - I_3 = \left[ \frac{2 \sin(4x)}{4} \right]_0^{\frac{\pi}{2}} = \frac{1}{2} (\sin(2\pi) - \sin(0)) = 0$.
Thus,$I_5 = I_3$.
Similarly,$I_3 - I_1 = \left[ \frac{2 \sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = [\sin(2x)]_0^{\frac{\pi}{2}} = \sin(\pi) - \sin(0) = 0$.
Thus,$I_3 = I_1$.
Finally,$I_1 = \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x} dx = \int\limits_0^{\frac{\pi}{2}} 1 dx = \frac{\pi}{2}$.
Hence,$I_5 = I_3 = I_1 = \frac{\pi}{2}$.

(D) We know that the signum function is defined as $\operatorname{sgn}(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases}$.
Also,note that $x \operatorname{sgn}(x) = |x|$.
Let $I = \int_{a}^{b} \operatorname{sgn}(x) \, dx$.
Consider the antiderivative of $\operatorname{sgn}(x)$,which is $f(x) = |x|$.
By the Fundamental Theorem of Calculus,$\int_{a}^{b} f'(x) \, dx = f(b) - f(a)$.
Since $\frac{d}{dx}(|x|) = \operatorname{sgn}(x)$ for $x \neq 0$,the integral is $\int_{a}^{b} \operatorname{sgn}(x) \, dx = [|x|]_{a}^{b} = |b| - |a|$.
Alternatively,since $|x| = x \operatorname{sgn}(x)$,we have $|b| - |a| = b \operatorname{sgn}(b) - a \operatorname{sgn}(a)$.
Therefore,both expressions $(A)$ and $(C)$ are equivalent.
Thus,the correct option is $(D)$.

(D) Let $I = \int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$.
Using partial fractions,we write $\frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}$.
Equating coefficients,we get $A(x^2+2x+2) + (Bx+C)(x+1) = 2x^2+3x+3$.
For $x = -1$,$A(1-2+2) = 2-3+3 \implies A = 2$.
Comparing coefficients of $x^2$,$A+B = 2 \implies B = 0$.
Comparing constants,$2A+C = 3 \implies 4+C = 3 \implies C = -1$.
So,$I = \int_{0}^{1} (\frac{2}{x+1} - \frac{1}{x^2+2x+2}) dx = \int_{0}^{1} (\frac{2}{x+1} - \frac{1}{(x+1)^2+1}) dx$.
$I = [2 \ln|x+1| - \tan^{-1}(x+1)]_{0}^{1} = (2 \ln 2 - \tan^{-1} 2) - (2 \ln 1 - \tan^{-1} 1) = 2 \ln 2 - \tan^{-1} 2 + \frac{\pi}{4}$.
Since $\tan^{-1} 2 + \cot^{-1} 2 = \frac{\pi}{2}$,we have $\frac{\pi}{4} - \tan^{-1} 2 = \frac{\pi}{4} - (\frac{\pi}{2} - \cot^{-1} 2) = \cot^{-1} 2 - \frac{\pi}{4}$.
Thus,$I = 2 \ln 2 + \cot^{-1} 2 - \frac{\pi}{4}$.
Both options $A$ and $B$ are equivalent. Therefore,the correct answer is $D$.

(D) The area $A$ under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ is given by the definite integral:
$A = \int_{0}^{h} e^{-x} dx$
To find the limit as $h \rightarrow \infty$,we evaluate:
$\lim_{h \rightarrow \infty} A = \lim_{h \rightarrow \infty} \int_{0}^{h} e^{-x} dx$
Integrating $e^{-x}$ gives $-e^{-x}$:
$= \lim_{h \rightarrow \infty} [-e^{-x}]_{0}^{h}$
Applying the limits of integration:
$= \lim_{h \rightarrow \infty} (-e^{-h} - (-e^{0}))$
$= \lim_{h \rightarrow \infty} (-e^{-h} + 1)$
Since $\lim_{h \rightarrow \infty} e^{-h} = 0$,we have:
$= -0 + 1 = 1$

(D) To find the area under the curve $y = xe^{-ax}$ from $x = 0$ to $x = \infty$,we evaluate the definite integral:
$A = \int_{0}^{\infty} xe^{-ax} dx$
Using integration by parts,where $u = x$ and $dv = e^{-ax} dx$,we have $du = dx$ and $v = -\frac{1}{a}e^{-ax}$:
$A = \left[ -\frac{x}{a}e^{-ax} \right]_{0}^{\infty} - \int_{0}^{\infty} \left( -\frac{1}{a}e^{-ax} \right) dx$
$A = (0 - 0) + \frac{1}{a} \int_{0}^{\infty} e^{-ax} dx$
$A = \frac{1}{a} \left[ -\frac{1}{a}e^{-ax} \right]_{0}^{\infty}$
$A = \frac{1}{a} \left( 0 - (-\frac{1}{a}) \right) = \frac{1}{a^2}$