Question

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

Answer 1

Let 1 + log x = t

∴ $\frac{1}{x}dx = dt$

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

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

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