How to solve $\int\frac{\ln x}{x^2(\ln (x)-1)^2}dx$ by substitution $$\int\frac{\ln x}{x^2(\ln (x)-1)^2}dx$$
Hello, I haven't be able to solve this integral, I've tried to do $u = \ln (x)-1$ but couldn't make it work, any insight?

$x\ln x-x=u\implies \ln x dx=du\implies\displaystyle \int\frac{du}{u^2}=\dfrac{u^{-1}}{-1}+C=\dfrac{1}{x-x\ln x}+C.$
