Question:
$\frac{(1+\log x)^{2}}{x}$
Solution:
Let $1+\log x=t$
$\therefore \frac{1}{x} d x=d t$
$\Rightarrow \int \frac{(1+\log x)^{2}}{x} d x=\int t^{2} d t$
$=\frac{t^{3}}{3}+\mathrm{C}$
$=\frac{(1+\log x)^{3}}{3}+\mathrm{C}$
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