## Calculus (3rd Edition)

$$\int \frac{ \ln(\ln x) \ln x}{x} d x=\frac{1}{2}(\ln x)^2\ln (\ln x)-\frac{1}{4} (\ln x)^2+c$$
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 $$\int \frac{ \ln(\ln x) \ln x}{x} d x=\frac{1}{2}(\ln x)^2\ln (\ln x)-\frac{1}{4} (\ln x)^2+c$$