Tích phân \(I = \int\limits_0^1 {\left( {2x + 1} \...

B

- Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = \left( {2x + 1} \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{1}{{x + 1}}dx\\v = {x^2} + x\end{array} \right.\).

\( \Rightarrow I = \left. {\left[ {\left( {{x^2} + x} \right)\ln \left( {x + 1} \right)} \right]} \right|_0^1 - \int\limits_0^1 {xdx} \\= \left. {\left[ {\left( {{x^2} + x} \right)\ln \left( {x + 1} \right)} \right]} \right|_0^1 - \left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1 \\= 2\ln 2 - \frac{1}{2}\)