Prove

$\frac{(x+1)(x+\log x)^{2}}{x}=\left(\frac{x+1}{x}\right)(x+\log x)^{2}=\left(1+\frac{1}{x}\right)(x+\log x)^{2}$

Let $(x+\log x)=t$

$\therefore\left(1+\frac{1}{x}\right) d x=d t$

$\Rightarrow \int\left(1+\frac{1}{x}\right)(x+\log x)^{2} d x=\int t^{2} d t$

$=\frac{t^{3}}{3}+\mathrm{C}$

$=\frac{1}{3}(x+\log x)^{3}+\mathrm{C}$