\(I=\int\left[\log (\log x)+\frac{1}{(\log x)^2}\right] d x\)
Let, \(\log x=t \Rightarrow d x=e^t d t\)
\(\therefore \quad I=\int e^t\left(\log (t)+\frac{1}{t^2}\right) d t\)
\(\begin{aligned}
& =\int e^t\left[\left(\log t+\frac{1}{t}\right)-\left(\frac{1}{t}-\frac{1}{t^2}\right)\right] d t \\
& =\int e^t\left(\log t+\frac{1}{t}\right) d t-\int e^t\left(\frac{1}{t}-\frac{1}{t^2}\right) d t
\end{aligned}\)
As, we know that,
\(\int e^x\left(f(x)+f^{\prime}(x)\right) d x=\int e^x f(x)+C \text {, }\)
\(\begin{aligned}
\text{So, } I & =e^t \log t-\frac{e^t}{t}+C \\
& =x \log (\log x)-\frac{x}{\log x}+C \quad\{\because t=\log x\} \\
& =x\left[\log (\log x)-\frac{1}{\log x}\right]+C
\end{aligned}\)