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Câu hỏi: Biết $\int\limits_{1}^{e}{\dfrac{\ln \text{x}}{{{(1+x)}^{2}}}d\text{x}}=\dfrac{a}{e+1}+b\ln \dfrac{2}{e+1}+c$, với $a,b,c\in \mathbb{Z}$. Tính $a+b+c$.
A. $-1$
B. 1
C. 3
D. 2
A. $-1$
B. 1
C. 3
D. 2
Đặt $\left\{ \begin{aligned}
& u=\ln \text{x} \\
& dv=\dfrac{d\text{x}}{{{(1+x)}^{2}}} \\
\end{aligned} \right. $ ta có $ \left\{ \begin{aligned}
& du=\dfrac{d\text{x}}{x} \\
& v=-\dfrac{1}{1+x} \\
\end{aligned} \right.$
Theo công thức tích phân từng phần có
$\int\limits_{1}^{e}{\dfrac{\ln \text{x}}{{{(1+x)}^{2}}}d\text{x}}=\left. -\dfrac{\ln \text{x}}{1+x} \right|_{1}^{e}+\int\limits_{1}^{e}{\dfrac{d\text{x}}{x(1+x)}}=\dfrac{-1}{e+1}+\int\limits_{1}^{e}{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)d\text{x}}$
$=\dfrac{-1}{e+1}+\left. \left( \ln \left| x \right|-\ln \left| x+1 \right| \right) \right|_{1}^{e}=\dfrac{-1}{e+1}+\ln \dfrac{2}{e+1}+1$.
Suy ra $a=-1,b=1,c=1$. Vậy $a+b+c=1$.
& u=\ln \text{x} \\
& dv=\dfrac{d\text{x}}{{{(1+x)}^{2}}} \\
\end{aligned} \right. $ ta có $ \left\{ \begin{aligned}
& du=\dfrac{d\text{x}}{x} \\
& v=-\dfrac{1}{1+x} \\
\end{aligned} \right.$
Theo công thức tích phân từng phần có
$\int\limits_{1}^{e}{\dfrac{\ln \text{x}}{{{(1+x)}^{2}}}d\text{x}}=\left. -\dfrac{\ln \text{x}}{1+x} \right|_{1}^{e}+\int\limits_{1}^{e}{\dfrac{d\text{x}}{x(1+x)}}=\dfrac{-1}{e+1}+\int\limits_{1}^{e}{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)d\text{x}}$
$=\dfrac{-1}{e+1}+\left. \left( \ln \left| x \right|-\ln \left| x+1 \right| \right) \right|_{1}^{e}=\dfrac{-1}{e+1}+\ln \dfrac{2}{e+1}+1$.
Suy ra $a=-1,b=1,c=1$. Vậy $a+b+c=1$.
Đáp án B.