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Câu hỏi: Biết $\int\limits_{0}^{2}{\dfrac{{{x}^{2}}{{e}^{2x}}-3{{e}^{x}}+2}{x{{e}^{x}}+1}}dx=a{{e}^{2}}+b+c\ln \left( 2{{e}^{2}}+1 \right)$ với $a,b,c$ là các số nguyên, giá trị của $T=a+{{b}^{2}}+{{c}^{3}}$ là
A. $T=27$
B. $T=0$
C. $T=-1$
D. $T=-2$
A. $T=27$
B. $T=0$
C. $T=-1$
D. $T=-2$
Ta có:
$\int\limits_{0}^{2}{\dfrac{{{x}^{2}}{{e}^{2x}}-3{{e}^{x}}+2}{x{{e}^{x}}+1}}dx=\int\limits_{0}^{2}{\dfrac{\left( {{x}^{2}}{{e}^{2x}}-1 \right)-3\left( {{e}^{x}}-1 \right)}{x{{e}^{x}}+1}}dx=\int\limits_{0}^{2}{\left( x{{e}^{x}}-1 \right)}dx-3\int\limits_{0}^{2}{\dfrac{{{e}^{x}}-1}{x{{e}^{x}}+1}}dx$
$=\left( x{{e}^{x}}-{{e}^{x}}-x \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.-3\int\limits_{0}^{2}{\dfrac{1-{{e}^{-x}}}{x+{{e}^{-x}}}dx}=\left( {{e}^{2}}-1 \right)-3\int\limits_{0}^{2}{\dfrac{1}{x+{{e}^{-x}}}d\left( x+{{e}^{-x}} \right)}$
$=\left( {{e}^{2}}-1 \right)-3\ln \left( x+{{e}^{-x}} \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.={{e}^{2}}+5-3\ln \left( 2{{e}^{2}}+1 \right)$
$\Rightarrow a=1;b=5;c=-3$. Kết luận: $T=-1$
$\int\limits_{0}^{2}{\dfrac{{{x}^{2}}{{e}^{2x}}-3{{e}^{x}}+2}{x{{e}^{x}}+1}}dx=\int\limits_{0}^{2}{\dfrac{\left( {{x}^{2}}{{e}^{2x}}-1 \right)-3\left( {{e}^{x}}-1 \right)}{x{{e}^{x}}+1}}dx=\int\limits_{0}^{2}{\left( x{{e}^{x}}-1 \right)}dx-3\int\limits_{0}^{2}{\dfrac{{{e}^{x}}-1}{x{{e}^{x}}+1}}dx$
$=\left( x{{e}^{x}}-{{e}^{x}}-x \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.-3\int\limits_{0}^{2}{\dfrac{1-{{e}^{-x}}}{x+{{e}^{-x}}}dx}=\left( {{e}^{2}}-1 \right)-3\int\limits_{0}^{2}{\dfrac{1}{x+{{e}^{-x}}}d\left( x+{{e}^{-x}} \right)}$
$=\left( {{e}^{2}}-1 \right)-3\ln \left( x+{{e}^{-x}} \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.={{e}^{2}}+5-3\ln \left( 2{{e}^{2}}+1 \right)$
$\Rightarrow a=1;b=5;c=-3$. Kết luận: $T=-1$
Đáp án C.