Question

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

Answer 1

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

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

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

and $dv=dx\Rightarrow v=x$

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

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

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

and $dv=dx\Rightarrow v=x$

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

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

$=x\log \log x-\frac{x}{\log x}+c$