Question

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

Answer 1

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

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

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

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

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

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

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