I know that the integral of $\int \frac{1}{x \ln x}\, \mathrm{d}x$ can easily be obtained through substitution for $u=\ln x$ with the result of $\ln \ln x+C$. My question is if this answer (or an equivalent one) can be obtained via integration by parts, and if not why?
I have tried the substitutions $u = \frac{1}{\ln x}$ and $dv = \frac{1}{x}\,dx$ which yeilds $$ \int \frac{1}{x \ln x} \, \mathrm{d}x = 1 + \int \frac{1}{x \ln x} \, \mathrm{d}x$$ which is no good.