Tích phân \(I = \int\limits_{ - 2}^2 {\left| {\fra...

A

- Hướng dẫn giải

\(I = \int\limits_{ - 2}^0 {\left| {\frac{{{x^2} - x - 2}}{{x - 1}}} \right|} dx\\ = - \int\limits_{ - 2}^{ - 1} {\left( {\frac{{{x^2} - x - 2}}{{x - 1}}} \right)} dx + \int\limits_{ - 1}^0 {\frac{{{x^2} - x - 2}}{{x - 1}}} dx\)

\(\begin{array}{l}{I_1} = - \int\limits_{ - 2}^{ - 1} {\left( {\frac{{{x^2} - x - 2}}{{x - 1}}} \right)} dx\\ = - \int\limits_{ - 2}^{ - 1} {\left( {x - \frac{2}{{x - 1}}} \right)} dx\\ = - \left. {\left( {\frac{{{x^2}}}{2} - 2\ln \left| {x - 1} \right|} \right)} \right|_{ - 2}^{ - 1}\\ = \frac{5}{2} + 2\ln 2 - 2\ln 3\end{array}\)

\({I_2} = \int\limits_{ - 1}^0 {\left( {\frac{{{x^2} - x - 2}}{{x - 1}}} \right)} dx = \left. {\left( {\frac{{{x^2}}}{2} - 2\ln \left| {x - 1} \right|} \right)} \right|_{ - 1}^0 = \frac{1}{2} - 2\ln 2\)

\(\Rightarrow I = {I_1} + {I_2} = 3 - 2\ln 3\)