Some limitations in Integrals.

first split the integral in 2: $\int_0^\infty \dfrac{\ln^2 (x)}{\sqrt{x} (1-x)^2} \text{dx} = \int_0^1 \dfrac{\ln^2 (x)}{\sqrt{x} (1-x)^2} \text{dx}+\int_1^\infty \dfrac{\ln^2 (x)}{\sqrt{x} (1-x)^2} \text{dx}\\= \int_0^1 \dfrac{\ln^2 (x)}{\sqrt{x} (1-x)^2} \text{dx}+ \int_0^1 \dfrac{\ln^2 (x)\sqrt{x}}{ (1-x)^2} \text{dx} = \int_0^1 \dfrac{\ln^2 (x)(1+x)}{\sqrt{x} (1-x)^2} \text{dx}$ the reason the integral was split in two was so that the bound would be from 0 to 1, allowing the use of the taylor series: $\dfrac{1}{(1-x)^2} = \sum_{n=1}^\infty n x^{n-1}, |x|\leq 1$ plugging this in the integral and interchanging summation and integral signs: $\sum_{n=1}^\infty n \int_0^1 \ln^2(x) x^{n-3/2}+\ln^2(x) x^{n-1/2} \text{dx}$ using $I(a) = \int_0^1 x^a \text{dx} = \dfrac{1}{a+1}\to I''(a) = \int_0^1 x^a \ln^2(x) \text{dx} = \dfrac{2}{(a+1)^3}\\ \int_0^1 \ln^2(x) x^{n-3/2}+\ln^2(x) x^{n-1/2} \text{dx} = \dfrac{2}{(n-1/2)^3}+\dfrac{2}{(n+1/2)^3}$ note differentiation under the integral sign was used. now the sum becomes $\sum_{n=1}^\infty \left(\dfrac{2n}{(n-1/2)^3}+\dfrac{2n}{(n+1/2)^3} \right)=\sum_{n=1}^\infty \left(\dfrac{2n}{(n-1/2)^3}+\dfrac{2(n-1)}{(n-1/2)^3} \right) =\sum_{n=1}^\infty \left(\dfrac{4n-2}{(n-1/2)^3}\right)\\=\sum_{n=1}^\infty \left(\dfrac{4}{(n-1/2)^2}\right)=\sum_{n=1}^\infty \left(\dfrac{16}{(2n-1)^2}\right)=16\zeta(2)(1-2^{-2}) =\boxed{2}\pi^2$

@Chew-Seong Cheong,@Jon Haussmann, please post your solution as I am tired trying this one.