Question

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

Answer 1

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

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

Integrating by parts, we get

Let $u=\log \left(\log x\right)\Rightarrow du=\frac{1}{x\log x}dx$

$dv=dx\Rightarrow v=x$

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

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

Let $u={\left(\log x\right)}^{-1}\Rightarrow du=-\frac{1}{x{\left(\log x\right)}^2}dx$

$dv=dx\Rightarrow v=x$

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

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

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

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