Đáp án đúng: A
Giải chi tiết:Ta có \(\dfrac{{x + 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{ - 2}}{{x - 1}} + \dfrac{3}{{x - 2}}\).
Do đó
\(\begin{array}{l}\int\limits_{}^{} {\dfrac{{x + 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}dx} = \int\limits_{}^{} {\left( {\dfrac{{ - 2}}{{x - 1}} + \dfrac{3}{{x - 2}}} \right)dx} = - 2\ln \left| {x - 1} \right| + 3\ln \left| {x - 2} \right| + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a\ln \left| {x - 1} \right| + b\ln \left| {x - 2} \right| + C \Rightarrow \left\{ \begin{array}{l}a = - 2\\b = 3\end{array} \right. \Rightarrow a + b = 1\end{array}\)
Chọn A.