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Câu hỏi: Biết $\int\limits_{2}^{e+1}{\dfrac{\ln \left( x-1 \right)}{{{\left( x-1 \right)}^{2}}}\text{d}x}=a+b{{e}^{-1}}\left( a,b\in \mathbb{R} \right)$, chọn khẳng định đúng trong các khẳng định sau:
A. $2{{a}^{2}}-3b=4$.
B. $2{{a}^{2}}-3b=8$.
C. $2{{a}^{2}}-3b=-4$.
D. $2{{a}^{2}}-3b=-8$.
A. $2{{a}^{2}}-3b=4$.
B. $2{{a}^{2}}-3b=8$.
C. $2{{a}^{2}}-3b=-4$.
D. $2{{a}^{2}}-3b=-8$.
Đặt $\left\{ \begin{aligned}
& u=\ln \left( x-1 \right) \\
& \text{d}v=\dfrac{1}{{{\left( x-1 \right)}^{2}}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x-1}\text{d}x \\
& v=-\dfrac{1}{x-1} \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{2}^{e+1}{\dfrac{\ln \left( x-1 \right)}{{{\left( x-1 \right)}^{2}}}\text{d}x}=\left. \left( -\dfrac{1}{x-1}\ln \left( x-1 \right) \right) \right|_{2}^{e+1}+\int\limits_{2}^{e+1}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}\text{d}x}$ $=-\dfrac{1}{e}-\left. \dfrac{1}{x-1} \right|_{2}^{e+1}$ $=-\dfrac{1}{e}-\dfrac{1}{e}+1$ $=-2{{e}^{-1}}+1$.
$\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=-2 \\
\end{aligned} \right.\Rightarrow 2{{a}^{2}}-3b=8$.
& u=\ln \left( x-1 \right) \\
& \text{d}v=\dfrac{1}{{{\left( x-1 \right)}^{2}}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x-1}\text{d}x \\
& v=-\dfrac{1}{x-1} \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{2}^{e+1}{\dfrac{\ln \left( x-1 \right)}{{{\left( x-1 \right)}^{2}}}\text{d}x}=\left. \left( -\dfrac{1}{x-1}\ln \left( x-1 \right) \right) \right|_{2}^{e+1}+\int\limits_{2}^{e+1}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}\text{d}x}$ $=-\dfrac{1}{e}-\left. \dfrac{1}{x-1} \right|_{2}^{e+1}$ $=-\dfrac{1}{e}-\dfrac{1}{e}+1$ $=-2{{e}^{-1}}+1$.
$\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=-2 \\
\end{aligned} \right.\Rightarrow 2{{a}^{2}}-3b=8$.
Đáp án B.