Question

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

Answer 1

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