Giả sử \(\int\limits_0^2 {\frac{{x - 1}}{{{x^2} + 4x + 3}}dx = a\ln 5 + b\ln 3;\,\,\,a,b \in R.} \) Tính P = ab

* Hướng dẫn giải

\(\begin{array}{l} \int\limits_0^2 {\frac{{x - 1}}{{{x^2} + 4x + 3}}dx = \int\limits_0^2 {\left( {\frac{2}{{x + 3}} - \frac{1}{{x + 1}}} \right)dx = \left( {2\ln \left| {x + 3} \right| - \ln \left| {x + 1} \right|} \right)\left| \begin{array}{l} ^2\\ _0 \end{array} \right. = 2\ln 5 - 3\ln 3} } \\ \Rightarrow \left\{ \begin{array}{l} a = 2\\ b = - 3 \end{array} \right. \Rightarrow ab = - 6 \end{array}\)