Find $\int \ln\ln x\,\mathrm dx$.
I tried$$\int \ln\ln x\,\mathrm dx=x\ln\ln x -\int \frac{1}{\ln x}\,\mathrm dx\\=x\ln x\ln x-\sqrt{x}{\ln x}-\int \frac{1}{(\ln x)^2}\,\mathrm dx$$ It seems more and more difficult. And I tried substitution, i.e. substituting the $\ln \ln x$ to $t$, then $=\int t\ e^{e^t}e^t\,\mathrm dt$ and then I can not integral it.
$$ \int\ln\ln(x)dx=x\ln\ln(x)-\int\frac{dx}{\ln x}=x\ln\ln(x)-\text{Li}(x)+c $$