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] &= \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{align} <\math>