Question

Evaluate:-
$\int \left[\log \left(\log x\right)+\frac{1}{{\left(\log x\right)}^2}\right]dx$

Answer 1

$\int \log \left(\log x\right)dx=\log \left(\log x\right)\int dx-\int \frac{d\left(\log \right(\log x\left)\right)}{dx}(\int dx)dx$ $(\because$ integrating by parts $)$

$=x\log \left(\log x\right)-\int \frac{1}{x(\log x{)}^2}\left(x\right)dx+c$

$=x\log \left(\log x\right)-\int \frac{1}{(\log x{)}^2}dx+c$

$⟹\int [\log \left(\log x\right)+\frac{1}{(\log x{)}^2}]dx=x\log \left(\log x\right)+c$