Tích phân \(I = \int\limits_0^1 {\left( {\frac{{ax...

A

- Hướng dẫn giải

\(\begin{array}{l}I = \int\limits_0^1 {\left( {\frac{{ax}}{{x + 1}} - 2ax} \right)dx} \\ = a\int\limits_0^1 {\frac{x}{{x + 1}}dx - 2a\int\limits_0^1 {xdx} } \\ = a\left. {\left( {x - \ln \left| {x + 1} \right|} \right)} \right|_0^1 - a\left. {\left( {{x^2}} \right)} \right|_0^1\\ = a\left( {1 - \ln 2} \right) - a\\ = - a\ln 2\end{array}\)