Question

Evaluate: $\int \log \left(\log x\right)+{\left(\log x\right)}^{-2}=?$

Answer 1

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

$=\int \log \underset{I}{\left(\log x\right)}.\underset{II}{1dx}+\int \frac{1}{{\left(\log x\right)}^2}dx$

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

$=\log \left(\log x\right)-\int \underset{I}{\frac{1}{\log x}}.\underset{II}{1dx}+\int \frac{1}{{\left(\log x\right)}^2}dx$

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

$=x\log \left(\log x\right)-\frac{x}{\log x}+c$