$$
\begin{aligned}
\int _{0}^{1}\frac {\sqrt x\ln x}{(1-x^{2})^{\frac {3}2}}dx
&=-\frac {\sqrt {2\pi }}4\Gamma ^{2}\left (\frac {3}4\right )\\
\int _{0}^{1}\frac {x^{\frac {1}2}\ln x\ln (1-x^2)}{(1-x^{2})^{\frac {3}2}}dx
&=\frac {\Gamma ^{2}\left (\frac {3}4\right )}{4\sqrt {2\pi }}\left (16G+\pi ^{2}-2\pi \log 2-4\pi \right )
\end{aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\sqrt x\ln x}{(1-x^{2})^{\frac {3}2}}dx
&=\frac {1}4\int _{0}^{1}x^{-\frac {1}4}(1-x)^{-\frac {3}2}\ln xdx\\
&=\frac {1}4\frac {d}{dt}\left. B\left (t,-\frac {1}2\right )\right|_{t=\frac {3}4}\\
&=\frac {1}4\frac {d}{dt}\left.\frac {\Gamma (t)\Gamma \left (-\frac {1}2\right )}{\Gamma \left (t-\frac {1}2\right )}\right|_{t=\frac {3}4}\\
&=\left.-\frac {\sqrt \pi}2 \frac {\Gamma '(t)\Gamma \left (t-\frac {1}2\right )-\Gamma (t)\Gamma '\left (t-\frac {1}2\right )}{\Gamma ^{2}\left (t-\frac {1}2\right )}\right|_{t=\frac {3}4}\\
&=\frac {\sqrt \pi }{2\Gamma ^{2}\left (\frac {1}4\right )}\left (\Gamma \left (\frac {3}4\right )\Gamma '\left (\frac {1}4\right )-\Gamma \left (\frac {1}4\right )\Gamma '\left (\frac {3}4\right )\right )\\
&=\frac {\sqrt \pi }2\Gamma ^{2}\left (\frac {3}4\right )\frac {\sin ^{2}\frac {\pi }4}{\pi ^{2}}\cdot \frac {\pi }{\sin \frac {\pi }4}\left (\psi \left (\frac {1}4\right )-\psi \left (\frac {3}4\right )\right )\\
&=\frac {1}2\sqrt {\frac {1}{2\pi} }\Gamma ^{2}\left (\frac {3}4\right )\left (\psi \left (\frac {1}4\right )-\psi \left (\frac {3}4\right )\right )\\
&=\textcolor {blue}{-\frac {\sqrt {2\pi }}4\Gamma ^{2}\left (\frac {3}4\right )}.
\end {aligned}
$$
関連するの積分(計算中)
$$
\begin {aligned}
\int _{0}^{\frac {\pi }2}\frac {x}{\sin ^{\frac {3}2}x}dx
&=\int _{0}^{\frac {\pi }2}\frac {x(2ie^{ix})^{\frac {3}2}}{(e^{2ix}-1)^{\frac {3}2}}dx\\
&=\int _{1}^{i}\frac {\frac {\ln z}i2(-1+i)z^{\frac {3}2}}{(z^{2}-1)^{\frac {3}2}}\frac {dz}{iz}\\
&=2\int _{1}^{0}\frac {x^{\frac {1}2}\ln x}{(1-x^{2})^{\frac {3}2}}dx+2\Re \int _{0}^{1}\frac {-i(\ln x+\frac {\pi i}2)(-1+i)x^{\frac {3}2}\frac {-1+i}{\sqrt 2}}{-i(1+x^2)^{\frac {3}2}ix}dx\\
&=-2\int _{0}^{1}\frac {x^{\frac {1}2}\ln x}{(1-x^2)^{\frac {3}2}}dx-2\sqrt 2\int _{0}^{1}\frac {x^{\frac {1}2}\ln x}{(1+x^{2})^{\frac {3}2}}dx\\
&=\sqrt {\frac {\pi }2}\Gamma ^{2}\left (\frac {3}4\right )-2\sqrt 2\int _{0}^{1}\frac {x^{\frac {1}2}\ln x}{(1+x^{2})^{\frac {3}2}}dx
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{\frac{\pi}2}\frac {x\ln \sin x}{\sin ^{\frac {3}2}x}dx
&=-2\int _{0}^{1}\frac {x^{\frac {1}2}\ln x\ln \frac {1-x^{2}}2}{(1-x^{2})^{\frac {3}2}}dx-2\sqrt 2\int _{0}^{1}\frac {x^{\frac {1}2}\ln x\ln \frac {1+x^{2}}2}{(1+x^2)^{\frac {3}2}}\\
\end {aligned}
$$
$$
\begin {aligned}
I_1&=\int _{0}^{1}\frac {x^{\frac {1}2}\ln x}{(1-x^{2})^{\frac {3}2}}dx\\
I_2&=\int _{0}^{1}\frac {x^{\frac {1}2}\ln x}{(1+x^{2})^{\frac {3}2}}dx\\
I_3&=\int _{0}^{1}\frac {x^{\frac {1}2}\ln x\ln (1-x^{2})}{(1-x^{2})^{\frac {3}2}}dx\\
I_4&=\int _{0}^{1}\frac {x^{\frac {1}2}\ln x\ln (1+x^{2})}{(1+x^{2})^{\frac {3}2}}dx
\end {aligned}
\\
\begin {aligned}
\int _{0}^{\frac {\pi }2}\frac {x(2-\ln \sin x)}{\sin ^{\frac {3}2}x}dx
&=-4I_1-4\sqrt 2I_2+2(I_3-I_1\ln 2)+2\sqrt 2(I_4-I_2\ln 2)\\
&=-2\left ((2+\ln 2)I_1-I_3\right )-2\sqrt 2((2+\ln 2)I_2-I_4)\\
&=-2\sqrt 2((2+\ln 2)I_2-I_4)-2\frac {\Gamma ^{2}\left (\frac {3}4\right )}{4\sqrt {2\pi }}\left (-(2+\ln 2)2\pi
- (16G+\pi ^{2}-2\pi \ln 2-4\pi )\right )\\
&=-2\sqrt 2\left ((2+\ln 2)I_2-I_4\right )+\frac {\Gamma ^{2}\left (\frac {3}4\right )}{2\sqrt {2\pi }}\left (16G+\pi ^{2}\right )
\end {aligned}
$$
$$
\begin {aligned}
2\sqrt 2\int _{0}^{1}\frac {x^{\frac {1}2}\ln x}{(1+x^{2})^{\frac {3}2}}dx
&=\frac {1}2\int _{0}^{1}(1-x^{2})^{-\frac {1}4}\left (\ln (1-x)-\ln (1+x)\right )dx\\
&=\int _{0}^{1}(1-x^{2})^{-\frac {1}4}\ln (1-x)dx-\frac {1}2\int _{0}^{1}(1-x^{2})^{-\frac {1}4}\ln (1-x^{2})dx\\
&=\int _{0}^{1}(1-x^{2})^{-\frac {1}4}\ln (1-x)dx+\frac {\Gamma ^{2}\left (\frac {3}4\right )}{\sqrt {2\pi }}(4-2\pi )
\end {aligned}
$$
$$
\begin {aligned}
2\sqrt 2\int _{0}^{1}\frac {x^{\frac {1}2}\ln x\ln \frac {1+x^{2}}2}{(1+x^{2})^{\frac {3}2}}dx
&=\frac {1}2\int _{0}^{1}(1-x^{2})^{-\frac {1}4}\left (\ln (1+x)-\ln (1-x)\right )\ln (1+x)dx
\end {aligned}
$$