Ta có: $I = \int\limits_1^2 {{x^5}} dx = \left. {\dfrac{{{x^6}}}{6}} \right|_1^2 = \dfrac{{21}}{2}$.

$\int\limits_{1}^{e}{\frac{f(x)}{x}dx}=\int\limits_{1}^{e}{f(x)d\ln x}=\left. f(x)\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\ln xf'(x)dx}=1$

$\Rightarrow f(e)-\int\limits_{1}^{e}{\ln xf'(x)dx}=1$

$\Leftrightarrow \int\limits_{1}^{e}{\ln xf'(x)dx}=f(e)-1=2-1=1$

Đáp án A: đúng theo tính chất tích phân.

Đáp án B: sai vì \(x\) không phải hằng số nên không đưa được ra ngoài dấu tích phân.

Đáp án C: đúng theo tính chất tích phân.

Đáp án D: đúng theo tính chất tích phân.

Đặt $u=x+1;dv=f'(x)dx$ thì $du=dx;v=f(x)$

Ta có:

$\begin{array}{l}\int\limits_0^1 {(x + 1)f'(x)d{\rm{x}} = 10}  \Leftrightarrow \left. {(x + 1)f(x)} \right|_0^1 - \int\limits_0^1 {f(x)d{\rm{x}} = 10 = } 2f(1) - f(0) - \int\limits_0^1 {f(x)d{\rm{x}}} \\ \to \int\limits_0^1 {f(x)d{\rm{x}}}  =  - 8.\end{array}$

Ta có $\int\limits_0^2 {g\left( x \right){\rm{d}}x}  = \int\limits_0^1 {g\left( x \right){\rm{d}}x}  + \int\limits_1^2 {g\left( x \right){\rm{d}}x}  = \int\limits_0^1 {g\left( x \right){\rm{d}}x}  + \int\limits_1^2 {g\left( t \right){\rm{d}}t} $

Suy ra $\int\limits_0^1 {g\left( x \right){\rm{d}}x}  = \int\limits_0^2 {g\left( x \right){\rm{d}}x}  - \int\limits_1^2 {g\left( t \right){\rm{d}}t}  =  - \,2 - 1 =  - \,3.$ 

Vậy $I = 2\,\int\limits_0^1 {f\left( x \right){\rm{d}}x}  - \int\limits_0^1 {g\left( x \right){\rm{d}}x}  = 2.4 - \left( { - \,3} \right) = 11.$

Đặt : \(\left\{ \begin{array}{l}u = f(x)\\dv = \sin {\rm{xdx}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = f'(x)dx\\v = - \cos x\end{array} \right.  \)

$\Rightarrow \int {f(x)\sin {\rm{x}}dx =  - f(x).\cos x + \int {f'(x).\cos xdx} }$

Nên suy ra \(f'(x) = {\pi ^x} \Rightarrow f(x) = \int {{\pi ^x}} dx = \dfrac{{{\pi ^x}}}{{\ln \pi }}\).

Đặt $t = 3x \Rightarrow dt = 3dx \Rightarrow dx = \dfrac{{dt}}{3}$, khi đó:

$\begin{array}{*{20}{l}}{\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x} {\rm{\;}} = \dfrac{1}{3}\int {f\left( t \right)dt} {\rm{\;}} = \dfrac{1}{3}\left( {2t\ln \left( {3t - 1} \right)} \right) + C}\\{ = \dfrac{1}{3}\left( {2.3x.\ln \left( {3.3x - 1} \right)} \right) + C = 2x\ln \left( {9x - 1} \right) + C}\end{array}$

Vậy $\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x} {\rm{\;}} = 2x\ln \left( {9x - 1} \right) + C$

\(\begin{align}  & f'\left( x \right)=-3.\frac{a}{{{\left( x+1 \right)}^{4}}}+b{{e}^{x}}+bx{{e}^{x}} \\  & \Rightarrow f'\left( 0 \right)=-3a+b=-22\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \\  & \int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( \frac{a}{{{\left( x+1 \right)}^{3}}}+bx{{e}^{x}} \right)dx}=a\int\limits_{0}^{1}{{{\left( x+1 \right)}^{-3}}dx}+b\int\limits_{0}^{1}{x{{e}^{x}}dx}=a{{I}_{1}}+b{{I}_{2}} \\  & {{I}_{1}}=\int\limits_{0}^{1}{{{\left( x+1 \right)}^{-3}}dx}=\left. \frac{{{\left( x+1 \right)}^{-2}}}{-2} \right|_{0}^{1}=\frac{-1}{2}\left( \frac{1}{4}-1 \right)=\frac{3}{8} \\ \end{align}\)

Đặt\(\left\{ \begin{array}{l}
u = x\\
dv = {e^x}dx
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
du = dx\\
v = {e^x}
\end{array} \right. \Rightarrow {I_2} = \left. {x{e^x}} \right|_0^1 - \int\limits_0^1 {{e^x}dx} = e - \left. {{e^x}} \right|_0^1 = e - \left( {e - 1} \right) = 1\)

\(\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\frac{3}{8}a+b=5\,\,\,\,\,\,\left( 2 \right)\) 

Từ (1) và (2) \(\Rightarrow \left\{ \begin{align}  & a=8 \\  & b=2 \\ \end{align} \right.\)

Dễ thấy các công thức a) đúng vì tích phân có hai cận bằng nhau thì có giá trị $0$.

Công thức c) là đúng theo tính chất tích phân.

Công thức b) sai vì \(\int\limits_a^b {f\left( x \right)dx}  =  - \int\limits_b^a {f\left( x \right)dx} \)

$\int {{x^2}\sqrt {4 + {x^3}} dx = \dfrac{1}{3}\int {\sqrt {4 + {x^3}} .d\left( {{x^3} + 4} \right)} } $$ = \dfrac{1}{3}\dfrac{{{{\left( {4 + {x^3}} \right)}^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} + C = \dfrac{2}{9}\sqrt {{{\left( {4 + {x^3}} \right)}^3}}  + C$.

Ta có: \(\int\limits_{ - 1}^0 {\left( {x + 1 + \dfrac{2}{{x - 1}}} \right)dx}  = \left. {\left( {\dfrac{{{x^2}}}{2} + x + 2\ln \left| {x - 1} \right|} \right)} \right|_{ - 1}^0 \)

$= \dfrac{1}{2} - 2\ln 2 \Rightarrow \left\{ \begin{array}{l}a = \dfrac{1}{2}\\b =  - 2\end{array} \right. \Rightarrow a + b =  - \dfrac{3}{2}$

Ta có \(\int\limits_{0}^{1}{\frac{\text{d}x}{\sqrt{x+1}+\sqrt{x}}}=\int\limits_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{{{\left( \sqrt{x+1} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}}\,\text{d}x}=\int\limits_{0}^{1}{\left( \sqrt{x+1}-\sqrt{x} \right)\,\text{d}x}=\left. \frac{2}{3}\left[ \sqrt{{{\left( x+1 \right)}^{3}}}-\sqrt{{{x}^{3}}} \right] \right|_{0}^{1}=\frac{4}{3}\left( \sqrt{2}-1 \right).\)

Mặt khác \(I=\frac{2}{3}\left( \sqrt{a}-b \right)=\frac{4}{3}\left( \sqrt{2}-1 \right)=\frac{2}{3}\left( \sqrt{8}-2 \right)\Rightarrow \left\{ \begin{align} & a=8 \\ & b=2 \\ \end{align} \right..\)

Vậy \(T=a+b=8+2=10.\)