Evaluate the following integrals:

Assume $\log (\log x)=t$

$d(\log (\log x))=d t$

(use chain rule to differentiate first)

$\Rightarrow \frac{1}{\text { xlogx }} \mathrm{dx}=\mathrm{dt}$

Put $\mathrm{t}$ and $\mathrm{dt}$ in given equation we get

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\ln |\mathrm{t}|+\mathrm{c}$

But $t=\log (\log (x))$

$=\ln |\log (\log (x))|+c$