7.1 Integration By Parts/37

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