Solution
= Solution
For $0<x<1$, expand $(1-x)^{-2}=\sum_{k\ge1}kx^{k-1}$. All terms after multiplication by $(\log x)^2$ are nonnegative, so <monotone convergence> permits integration term by term. Substituting $x=e^{-t}$ and integrating by parts twice gives
$$
\int_0^1x^{k-1}(\log x)^2dx=\int_0^\infty e^{-kt}t^2dt=\frac2{k^3}.
$$
Consequently the integral is $\sum_{k\ge1}2/k^2$, and the supplied sum gives $\boxed{\pi^2/3}$.