Solving $\int_0^1(1-u)^{(n-i)} u^i \log(-\log(1-u))^2 du$?
As @ThomasAndrews mentions in his comment, you can obtain this case from the linked answer as follows: $$ \int_0^1 (1-u)^{(n-i)} u^i \ln^2(-\ln(1-u)) \mathrm{d}u =\sum_{j=0}^{i}\binom{i}{j} (-1)^j \int_{0}^{1} (1-u)^{n-i+j} \ln^2(-\ln(1-u)) \mathrm{d}u $$ and after some direct simplifications, this results in $$ \boxed{\int_0^1 (1-u)^{(n-i)} u^i \ln^2(-\ln(1-u)) \mathrm{d}u = \frac{\pi^2}{6} \frac{i! (n-i)!}{(n+1)!} +\sum_{k=1}^{i+1} \binom{i}{k-1} (-1)^{k+1} \frac{\left(\gamma + \ln(n-i+k) \right)^2}{(n-i+k)}} $$ which I doubt can be simplified further.