Question:
Evaluate: $\int \frac{(1+\log x)^{2}}{x} d x$.
Solution:
let $1+\log x=t$
Differentiating on both sides we get,
$\frac{1}{x} d x=d t$
Substituting it in $\int \frac{(1+\log x)^{2}}{x}$ we get,
$=\int t^{2} d t$
$=\frac{t^{3}}{3}+c$
$=\frac{(1+\log x)^{3}}{3}+c$