Question

Integrate the function $\frac{(1+\log x{)}^2}{x}$

Answer 1

Put $1+\log x=t$
$\therefore \frac{1}{x}dx=dt$
$\Rightarrow \int \frac{(1+\log x{)}^2}{x}dx=\int t^{2}dt$
$=\frac{t^3}{3}+C$
$=\frac{(1+\log x{)}^3}{3}+C$