Solving integral of (1-u)^n log(-log(1-u)) du?
By the substitution $x = \ln\left(\frac{1}{1-u} \right)$ your integral is equal to $$ \int_{0}^{\infty}e^{-x(n+1)} \ln^2(x) \, \mathrm{d}x $$ which by this answer gives you $$ \boxed{\int_0^1 (1-u)^n \ln^2\left(\ln\left(\frac{1}{1-u}\right)\right)\, \mathrm{d}u = \frac{1}{n+1}\left(\dfrac{\pi^2}{6} + \left(\gamma+ \ln(n+1)\right)^2\right)} $$