7.1 Integration By Parts/37

$\int x\ln({x})dx$
${\begin{aligned}u&=x+1\\[2ex]x&=u-1\\[2ex]du&=dx\\[2ex]\end{aligned}}$
$\int (u-1)\ln({u})du$

$w=\ln {u}\qquad dv=(u-1)dv$

$du={\frac {1}{u}}du\qquad v={\frac {1}{2}}u^{2}-u$

${\begin{aligned}\int (u-1)\ln({u})du&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u^{2}-u)\cdot {\frac {1}{u}}du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u-1)du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-[{\frac {1}{4}}u^{2}-u]+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}(x^{2}+2x+1)-(1+x))-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}x^{2}+x+{\frac {1}{2}}-x-1)-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}-{\frac {1}{2}}x-{\frac {1}{4}}+1+x+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}+{\frac {1}{2}}x+{\frac {3}{4}}+c\\[2ex]\end{aligned}}$