\( \smallint x\ln \left( {x + 1} \right)dx\) bằng
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Lời giải:
Báo saiĐặt \(\left\{ \begin{array}{l} u = \ln (x + 1)\\ dv = xdx \end{array} \right. \to \left\{ \begin{array}{l} du = \frac{1}{{x + 1}}dx\\ v = \frac{{{x^2}}}{2} \end{array} \right.\)
Khi đó
\(\begin{array}{l} \smallint x\ln \left( {x + 1} \right)dx = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{1}{2}\smallint \frac{{{x^2}}}{{x + 1}}dx\\ = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{1}{2}\smallint \left( {x - 1 + \frac{1}{{x + 1}}} \right)dx = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{{{x^2}}}{4} + \frac{1}{2}x - \frac{1}{2}\ln \left( {x + 1} \right) + C\\ = \left( {\frac{{{x^2}}}{2} - \frac{1}{2}} \right)\ln \left( {x + 1} \right) - \frac{1}{4}\left( {{x^2} - 2x + 1} \right) + C'\\ = \left( {\frac{{{x^2}}}{2} - \frac{1}{2}} \right)\ln \left( {x + 1} \right) - \frac{1}{4}{\left( {x - 1} \right)^2} + C' \end{array}\)
