Question

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

Answer 1

$\begin{array}{c}\int \left(\log \left(\log x\right)+\frac{1}{{\left(\log \right)}^2}\right)dx\\ =\int \log \left(logx\right)dx+\int \frac{1}{{\left(\log x\right)}^2}dx\\ =x\log \left(\log x\right)-\int \frac{1}{x\log x}\cdot xdx+\int \frac{1}{{\left(\log x\right)}^2}dx\\ =x\log \left(\log x\right)-\left\{\frac{x}{\log x}-\int -\frac{1}{{\left(\log x\right)}^2}dx\right\}+\int \frac{1}{{\left(\log x\right)}^2}dx\\ =x\log \left(\log x\right)-\frac{x}{\log x}+C\\ Hence,solved.\end{array}$