7.1 Integration By Parts/37

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int x\ln({x})dx }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= x + 1 \\[2ex] x &= u-1 \\[2ex] du &= dx \\[2ex] \end{align} }
${\begin{aligned}\int (u-1)\ln({u})du&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u^{2}-u)\cdot {\frac {1}{u}}du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u-1)du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-[{\frac {1}{4}}u^{2}-u]+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}(x^{2}+2x+1)-(1+x))-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}x^{2}+x+{\frac {1}{2}}-x-1)-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}-{\frac {1}{2}}x-{\frac {1}{4}}+1+x+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}+{\frac {1}{2}}x+{\frac {3}{4}}+c\\[2ex]\end{aligned}}$