The antiderivative of $f\left(x\right)=\log (\log x)+$$(\log x{)}^{-2}$ 

whose graph passes through $(e,e)$ is

 

An antiderivative of $f \left(x\right) = F\left(x\right) =$

$\begin{array}{c}\int \left(\log (\log x)+(\log x{)}^{-2}\right)\mathrm{d}x+C\\ =x\log (\log x)-\int \frac{x}{x\log x}\mathrm{d}x+\int (\log x{)}^{-2}\mathrm{d}x+C\end{array}$

$\text{ (integrating by parts the first term) }$

$\begin{array}{l}=x\log (\log x)-\left[x(\log x{)}^{-1}+\int (\log x{)}^{-2}\mathrm{d}x\right]+\\ \int (\log x{)}^{-2}\mathrm{d}x+C\end{array}$

$\begin{array}{r}\text{ (again integrating by parts }\left.(\log x{)}^{-1}\right)\\ =x\log (\log x)-x(\log x{)}^{-1}+C\end{array}$

Putting $x=e$ we have $e=0-e+C\text{ so }C=2e$

Thus $F\left(x\right)=x\left(\log (\log x)-(\log x{)}^{-1}\right)+2e$