複素積分で実積分を解く。

\begin{align} I&=\int_{0}^{\infty}\frac{\log^2x}{\sqrt{x}(1-x)^2}dx \end{align}
$$f(z)=\frac{\log^2x}{\sqrt{z}(1-z)^2}$$
積分経路
$$C=C_1+C_2+C_3+C_4$$
$$0=C_1+C_3$$
\begin{align} C_1&=\int_{\ips}^{R}f(x)dx \\&=I\qquad(\ips\to0,R\to\infty) \end{align}
\begin{align} f(-z)&=\frac{\log^2(-z)}{\sqrt{-z}(1+z)^2} \\&=\frac{(\log z+i\pi)^2}{i\sqrt{z}(1+z)^2} \\&=-i\frac{\log^2z}{\sqrt{z}(1+z)^2}+2\pi\frac{\log z}{\sqrt{z}(1+z)^2}+i\pi\frac{1}{\sqrt{z}(1+z)^2} \end{align}
\begin{align} C_2&=\int_{-R}^{-\ips}f(x)dx \\&=\int_{\ips}^{R}f(-x)dx \\&=2\pi\int_{0}^{\infty}\frac{\log x}{\sqrt{x}(1+x)^2}dx+i\left(-\int_{0}^{\infty}\frac{\log^2x}{\sqrt{x}(1+x)^2}+\pi^2\int_{0}^{\infty}\frac{dx}{\sqrt{x}(1+x)^2}\right)\qquad(\ips\to0,R\to\infty) \\&=2\pi I_1+ai\qquad(a\in\mathbb{R}) \end{align}
\begin{align} I_1&=\int_{0}^{\infty}\frac{\log x}{\sqrt{x}(1+x)^2}dx \\&=4\int_{0}^{\infty}\frac{\log x}{(1+x^2)^2}dx \end{align}
$$f(z)=\frac{\log z}{(1+z^2)^2}$$
積分経路
$$C=C_1+C_2+C_3+C_4$$
$$C=C_1+C_3$$
\begin{align} C_1&=\int_{\ips}^{R}f(x)dx \\&=\frac{1}{4}I_1\qquad(\ips\to0,R\to\infty) \end{align}
\begin{align} f(-z)&=\frac{\log z+i\pi}{(1+z^2)^2} \\&=f(z)+i\pi\frac{1}{(1+z^2)^2} \end{align}
\begin{align} C_2&=\int_{-R}^{-\ips}f(x)dx \\&=\int_{\ips}^{R}f(-x)dx \\&=\int_{\ips}^{R}f(x)dx+i\pi\int_{\ips}^{R}\frac{dx}{(1+x^2)^2} \\&=\frac{1}{4}I_1+bi\qquad(b\in\mathbb{R})(\ips\to0,R\to\infty) \end{align}
$$0=I+2\pi I_1+ai$$
$$I=-2\pi I_1$$
\begin{align} C&=2\pi i\underset{z=i}{\mathrm{Res}}f(z) \\&=2\pi i\lim_{z\to i}\frac{d}{dz}(z-i)^2\frac{\log z}{(z+i)^2} \\&=2\pi i\lim_{z\to i}\frac{\log z}{(z+i)^2} \\&=2\pi i\lim_{z\to i}\frac{i+z-2z\log z}{z(z+i)^3} \\&=2\pi i\frac{i+i-2i\log i}{i(i+i)^3} \\&=-\frac{\pi}2-\frac{\pi}{2}i \end{align}
$$-\frac{\pi}2-\frac{\pi}{2}i=\frac{1}4I_1+\frac{1}4I_1+bi$$
$$I_1=-\pi$$
$$I=-2\pi I_1$$
$$I=2\pi^2$$