+) $\int\limits_1^2 {{e^x}dx} = \left. {{e^x}} \right|_1^2 = {e^2} - e$

+) \(\int\limits_0^1 {2dx}  = \left. {2x} \right|_0^1 = 2\),

+) \(\int\limits_0^1 {xdx}  = \left. {\dfrac{{{x^2}}}{2}} \right|_0^1 = \dfrac{1}{2}\)

+) \(\int\limits_0^{\dfrac{\pi }{2}} {\sin xdx}  = \left. { - \cos x} \right|_0^{\dfrac{\pi }{2}} = 1\)

Vậy chỉ có đáp án B là có tích phân bằng \(2\).

Ta có \(\int{\dfrac{1}{x}\,\text{d}x}=\ln \left| x \right|+C\ne \ln x+C.\)

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

\(\int\limits_{1}^{2}{{{(x+3)}^{2}}dx}=\frac{1}{3}\left. {{(x+3)}^{3}} \right|_{1}^{2}=\frac{1}{3}\left( {{5}^{3}}-{{4}^{3}} \right)=\frac{61}{3}\)

Các mệnh đề A, B, C đều đúng. Mệnh đề D sai.

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

\(\left( {\sin x} \right)' = \cos x \Rightarrow y = \sin x\) là một nguyên hàm của hàm số $y = \cos x$.

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

Ta có: \(t = {x^2} \Rightarrow dt = 2xdx \Rightarrow xdx = \dfrac{{dt}}{2} \)

$\Rightarrow xf\left( {{x^2}} \right)dx = f\left( {{x^2}} \right).xdx = f\left( t \right).\dfrac{{dt}}{2} = \dfrac{1}{2}f\left( t \right)dt$

$\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 \(I=\int\limits_{1}^{2}{\frac{\text{d}x}{{{x}^{5}}+{{x}^{3}}}}=\int\limits_{1}^{2}{\frac{\text{d}x}{{{x}^{3}}\left( {{x}^{2}}+1 \right)}}=\frac{1}{2}\int\limits_{1}^{2}{\frac{2x\,\text{d}x}{{{x}^{4}}\left( {{x}^{2}}+1 \right)}}=\frac{1}{2}\int\limits_{1}^{2}{\frac{\text{d}\left( {{x}^{2}} \right)}{{{x}^{4}}\left( {{x}^{2}}+1 \right)}}=\frac{1}{2}\int\limits_{1}^{4}{\frac{\text{d}t}{{{t}^{2}}\left( t+1 \right)}}\)

Xét \(\int\limits_{1}^{4}{\frac{\text{d}t}{{{t}^{2}}\left( t+1 \right)}}=\int\limits_{1}^{4}{\frac{t+1-t}{{{t}^{2}}\left( t+1 \right)}\,\text{d}t}=\int\limits_{1}^{4}{\left( \frac{1}{{{t}^{2}}}-\frac{1}{t\left( t+1 \right)} \right)\,\text{d}t}=\int\limits_{1}^{4}{\frac{\text{d}t}{{{t}^{2}}}}-\int\limits_{1}^{4}{\frac{\text{d}t}{t\left( t+1 \right)}}\)

\(\begin{align}  & =\left. -\frac{1}{t} \right|_{1}^{4}-\int\limits_{1}^{4}{\left( \frac{1}{t}-\frac{1}{t+1} \right)dt=\frac{3}{4}-\left. \left( \ln t-\ln \left( t+1 \right) \right) \right|_{1}^{4}} \\ & =\frac{3}{4}-\ln 4+\ln 5-\ln 2=\frac{3}{4}-3\ln 2+\ln 5. \\\end{align}\)

Khi đó \(I=\frac{1}{2}\left( \frac{3}{4}-3\ln 2+\ln 5 \right)=\frac{1}{2}.\ln 5-\frac{3}{2}.\ln 2+\frac{3}{8}.\) Vậy \(a+2b+4c=-\,1.\)

Đặt \({{x}^{2}}+9=t\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{1}{2}dt\).

Đổi cận:

$\begin{array}{l}
x = 0 \Rightarrow t = 9\\
x = 4 \Rightarrow t = 25
\end{array}$

Khi đó, ta có: \(I=\int\limits_{0}^{4}{x\ln ({{x}^{2}}+9)dx=}\frac{1}{2}\int\limits_{9}^{25}{\ln tdt}=\frac{1}{2}\left[ \left. t.\ln \left| t \right| \right|_{9}^{25}-\int_{9}^{25}{td(\ln t)} \right]=\frac{1}{2}\left[ t.\ln \left. t \right|_{9}^{25}-\int_{9}^{25}{t.\frac{1}{t}dt} \right]\)

\(=\frac{1}{2}\left[ t.\ln \left. t \right|_{9}^{25}-\int_{9}^{25}{dt} \right]=\frac{1}{2}\left[ t.\ln \left. t \right|_{9}^{25}-\left. t \right|_{9}^{25} \right]=\frac{1}{2}\left[ \left( 25\ln 25-9\ln 9 \right)-(25-9) \right]=25\ln 5-9\ln 3-8\)

Suy ra, \(a=25,\,b=-9,\,c=-8\Rightarrow T=a+b+c=8\)