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

$$\begin{gathered} I=\int \left[\log \left(\log x\right)+(\log x{)}^{-2}\right]dx \\ substitute \log x=t\Rightarrow x=e^t\Rightarrow dx=e^{t}dt \\ \Rightarrow I=\int \left(\log t+\frac{1}{t^{ 2}}\right)e^{t}dt \\ \Rightarrow I=\int \left(\left(\log t-\frac{1}{t}\right)+\left(\frac{1}{t}+\frac{1}{t^{ 2}}\right)\right)e^{t}dt \\ \Rightarrow I=e^t\left(\log t-\frac{1}{t}\right)+C \\ \Rightarrow I=x\left\{\log \left(\log x\right)-\frac{1}{\log x}\right\}+c \end{gathered}$$