Let $u=\ln x$. Then $d u=\frac{1}{x} d x$. Thus we have
$$
\begin{aligned}
\int \frac{\ln \ln x}{x \ln x} d x &=\int \frac{\ln u}{u} d u \text { Let } v=\ln u, \text { then } d v=\frac{1}{u} d u \\
&=\int v d v \\
&=\frac{1}{2} v^{2}+C \\
&=\frac{1}{2}(\ln u)^{2}+C \\
&=\frac{1}{2}(\ln \ln x)^{2}+C
\end{aligned}
$$
