Câu hỏi 25

Đặt  \(\left\{ \begin{array}{l}\ln \left( {x - 1} \right) = u\\\dfrac{1}{{{{\left( {x - 1} \right)}^2}}}dx = dv\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{1}{{x - 1}}dx = du\\ - \dfrac{1}{{x - 1}} = v\end{array} \right.\)

 Ta có \(\int\limits_2^{e + 1} {\dfrac{{\ln \left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}dx = \ln \left( {x - 1} \right).\left. {\left( { - \dfrac{1}{{x - 1}}} \right)} \right|} _2^{e + 1} + \int\limits_2^{e + 1} {\dfrac{1}{{{{\left( {x - 1} \right)}^2}}}dx} \)

                               \( =  - \dfrac{1}{e} - \left. {\dfrac{1}{{x - 1}}} \right|_2^{e + 1} =  - \dfrac{1}{e} - \dfrac{1}{e} + 1 = 1 - 2.{e^{ - 1}}\)

Suy ra \(a = 1;\,\,\,b =  - 2 \Rightarrow a + b =  - 1.\)

Chọn B.