Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = xdx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{{x + 1}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.\)
Khi đó \(\int {x\ln \left( {x + 1} \right)dx} \)\( = \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{x + 1}}dx} \) \( = \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\left( {x - 1 + \dfrac{1}{{x + 1}}} \right)dx} \)
\( = \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{{{x^2}}}{4} + \dfrac{1}{2}x - \dfrac{1}{2}\ln \left( {x + 1} \right) + C\) \( = \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left( {x + 1} \right) - \dfrac{1}{4}\left( {{x^2} - 2x + 1} \right) + C'\)
\( = \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left( {x + 1} \right) - \dfrac{1}{4}{\left( {x - 1} \right)^2} + C'\)
Chọn C.
