Compute Integral $\int _{0}^{\infty} \frac {(\ln x)^2}{\sqrt x (1-x)^2} \mathrm dx$

Call this integral $I$. The identity $\int_0^\infty f(x)dx=\int_0^1\left(f(x)+\frac{1}{x^2}f(\frac{1}{x})\right)\mathrm{d}x$ gives $$I=\int_0^1\frac{x^{1/2}+x^{-1/2}}{(1-x)^2}\ln^2 x\mathrm{d}x.$$We can expand $(1-x)^{-2}$ as a power series $\sum_{k\ge 1}kx^{k-1}$, and $$\int_0^1 x^k\ln^2 x\mathrm{d}x=\frac{2}{(k+1)^3},$$ so $$\int_0^1 x^{k-1}(x^{1/2}+x^{-1/2})\ln^2 x\mathrm{d}x=\frac{2}{(k+1/2)^3}+\frac{2}{(k-1/2)^3}=\frac{64k(4k^2+3)}{(4k^2-1)^3}.$$Hence $$I=64\sum_{k\ge 1}\frac{k^2(4k^2+3)}{(4k^2-1)^3}\\=8\sum_{k\ge 1}\left(\frac{1}{(2k-1)^3}-\frac{1}{(2k+1)^3}+\frac{1}{(2k-1)^2}+\frac{1}{(2k+1)^2}\right)=8\left(1+2\sum_{k\ge 1}\frac{1}{(2k-1)^2}-1\right)\\=16\times\frac{\pi^2}{8}=2\pi^2.$$

Let $t=\sqrt{x}$, then $$\int _{0}^{\infty} \frac {(\ln x)^2}{\sqrt x (1-x)^2} \,dx=\int _{0}^{\infty} \frac {(\ln t^2)^2}{t (1-t^2)^2} \,2tdt= 8\int _{0}^{\infty} \frac {(\ln t)^2}{(1-t^2)^2} \,dt= \frac{8\pi^2}{4}=2\pi^2$$ where at the last step we use Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$ suggested by mrtaurho.

As regards $\int_0^\infty\frac{\ln^2x}{(1-x^2)^2}\,dx$, this is a variant of TheSimpliFire's approach given in a comment in the linked question.

By letting $x=e^t$ we get $$\begin{align*} \int_0^\infty\frac{\ln^2x}{(1-x^2)^2}\,dx &=\int_{-\infty}^\infty\frac{t^2e^{t}}{(1-e^{2t})^2}\,dt\\ &=\int_{0}^{+\infty}\frac{t^2e^{-t}}{(1-e^{-2t})^2}\,dt+\int_{0}^\infty\frac{t^2e^{-3t}}{(1-e^{-2t})^2}\,dt\\ &=\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+1)t}\,dt+\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+3)t}\,dt\\ &=\sum_{n=0}^\infty\frac{2(1+n)}{(2n+1)^3}+\sum_{n=0}^\infty\frac{2(1+n)}{(2n+3)^3}\\ &=\frac{\pi^2+7\zeta(3)}8+\frac{\pi^2-7\zeta(3)}8=\frac{\pi^2}{4}. \end{align*}$$