Answer
$$\int \frac{ \ln(\ln x) \ln x}{x} d x=\frac{1}{2}(\ln x)^2\ln (\ln x)-\frac{1}{4} (\ln x)^2+c$$
Work Step by Step
Let $u=\ln x$, then $du=\frac{1}{ x}dx$. Hence, we have
$$\int \frac{ \ln(\ln x) \ln x}{x} d x= \int u \ln u \ du
.$$
Now we rewrite with the variable z to avoid confusion with variable names and use integration by parts as follows; let $u\ln z$ and $dv=u$:
$$ \int z \ln z \ dz=\int u dv= uv-\int vdu\\
=\frac{1}{2}z^2\ln z-\frac{1}{2}\int z dz\\
=\frac{1}{2}z^2\ln z-\frac{1}{4} z^2 .$$
Hence $$\int \frac{ \ln(\ln x) \ln x}{x} d x=\frac{1}{2}(\ln x)^2\ln (\ln x)-\frac{1}{4} (\ln x)^2+c$$
