Đáp án đúng: A
Giải chi tiết:Đặt \(I = \int\limits_2^3 {\ln \left( {{x^2} - x} \right)dx} \).
Đặt \(\left\{ \begin{array}{l}u = \ln \left( {{x^2} - x} \right)\\dv = dx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = \dfrac{{2x - 1}}{{{x^2} - x}}dx\\v = x\end{array} \right.\).
\(\begin{array}{l} \Rightarrow I = \left. {x\ln \left( {{x^2} - x} \right)} \right|_2^3 - \int\limits_2^3 {\dfrac{{2x - 1}}{{x - 1}}dx} \\\,\,\,\,\,\,\,\,\,\, = 3\ln 6 - 2\ln 2 - \int\limits_2^3 {\left( {2 + \dfrac{1}{{x - 1}}} \right)dx} \\\,\,\,\,\,\,\,\,\,\, = 3\ln \left( {3.2} \right) - 2\ln 2 - \left. {\left( {2x + \ln \left| {x - 1} \right|} \right)} \right|_2^3\\\,\,\,\,\,\,\,\,\,\, = 3\ln 3 + 3\ln 2 - 2\ln 2 - \left( {6 + \ln 2 - 4} \right)\\\,\,\,\,\,\,\,\,\,\, = 3\ln 3 + \ln 2 - 2 - \ln 2 = 3\ln 3 - 2\end{array}\)
\( \Rightarrow 3\ln 3 - 2 = a\ln 3 - b \Rightarrow \left\{ \begin{array}{l}a = 3\\b = 2\end{array} \right. \Rightarrow a - b = 3 - 2 = 1\).
Chọn A.
