= Solution
For $0<x<1$, expand $(1-x)^{-2}=\sum_{k\geq1}kx^{k-1}$. Multiplying by $(\log x)^2$ gives nonnegative terms, so the <monotone convergence theorem> permits termwise integration. Put $x=e^{-t}$ to obtain
$$
\int_0^1x^{k-1}(\log x)^2\,dx
=\int_0^\infty t^2e^{-kt}\,dt=\frac2{k^3},
$$
where scaling followed by two integrations by parts evaluates the last integral. Hence
$$
\boxed{\int_0^1\left(\frac{\log x}{1-x}\right)^2dx
=2\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}{3}.}
$$
The finite value also establishes integrability at both endpoints.
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