Tích phân \(I = \int\limits_0^1 {\frac{1}{{{x^2} - x - 2}}} {\rm{d}}x\) có giá trị bằng.

* Hướng dẫn giải

\(\begin{array}{l} I = \int\limits_0^1 {\frac{1}{{{x^2} - x - 2}}} {\rm{d}}x\\ = \int\limits_0^1 {\frac{1}{{\left( {x + 1} \right)\left( {x - 2} \right)}}} {\rm{d}}x\\ = \frac{1}{3}\left( {\int\limits_0^1 {\frac{1}{{x - 2}}} {\rm{d}}x - \int\limits_0^1 {\frac{1}{{x + 1}}} {\rm{d}}x} \right)\\ = \frac{1}{3}\left. {\left( {\ln \left| {x - 2} \right| - \ln \left| {x + 1} \right|} \right)} \right|_0^1\\ = - \frac{{2\ln 2}}{3} \end{array}\)