Prove

Let $1+\log x=t$

$\therefore \frac{1}{x} d x=d t$

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

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

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