Closed form for $\int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx$ This is somewhat similar to my previous question: Closed form for $\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx$
Is it possible to find a closed form for this integral?
$$Q=\int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx$$
 A: First, we transform the integral into a more computable form by using some substitutions.
$$\begin{align*}\displaystyle Q &= \int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx\\ &=\int_0^1 \sqrt{\frac{1+u}{u(1-u)}}\log \left( \frac{(1+u)(1-u)}{u}\right)du \quad \color{blue}{\text{where }u=1-x}\\
&= \int_0^1 \frac{1+u}{\sqrt{u(1-u^2)}}\log \left( \frac{1-u^2}{u}\right)du \\
&= \frac{1}{2}\int_0^1 \frac{1+\sqrt{t}}{t^{\frac{3}{4}}\sqrt{1-t}}\log \left(\frac{1-t}{\sqrt{t}} \right)dt \quad \color{blue}{\text{where }t=u^2} \\
&= \frac{1}{2}\int_0^1 \frac{\log(1-t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt-\frac{1}{4}\int_0^1 \frac{\log(t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt \\
&\quad +\frac{1}{2}\int_0^1 \frac{\log(1-t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt-\frac{1}{4}\int_0^1 \frac{\log(t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt \tag{1}
\end{align*}$$
These four integrals can be evaluated by calculating derivatives of beta function in terms of digamma function. For e.g.
$$
\begin{align*}
\int_0^1 \frac{\log(1-t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt  &= \frac{d}{dz}\left\{ \int_0^1 t^{-\frac{3}{4}}(1-t)^{z-1} \; dt\right\}_{z=\frac{1}{2}}\\&= \frac{d}{dz}\left\{ \frac{\Gamma \left( \frac{1}{4}\right)\Gamma(z)}{\Gamma \left( \frac{1}{4}+z\right)} \right\}_{z=\frac{1}{2}}\\ &= \frac{\Gamma \left( \frac{1}{4}\right)\sqrt{\pi}}{\Gamma \left( \frac{3}{4}\right)}\left\{\psi_0 \left(\frac{1}{2} \right) -\psi_0 \left(\frac{3}{4} \right)\right\}\\ &= \pi^{3/2}\frac{\sqrt{2}}{\Gamma \left( \frac{3}{4}\right)^2}\left\{\log 2-\frac{\pi}{2} \right\} \tag{2} \\
\end{align*}
$$
To get the last expression, I used the special values
$$
\begin{align*}
\psi_0 \left(\frac{3}{4}\right) &= -\gamma +\frac{\pi}{2}-3\log 2 \\
\psi_0 \left(\frac{1}{2}\right) &= -\gamma -2\log 2
\end{align*}
$$
Using the same technique, the other three integrals can be evaluated:
$$
\begin{align*}
\int_0^1 \frac{\log(t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt&= -\pi^{5/2}\frac{\sqrt2}{\Gamma \left( \frac{3}{4}\right)^2}\tag{3} \\
\int_0^1 \frac{\log(t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt&=\frac{(4\pi-16)\Gamma \left( \frac{3}{4}\right)^2}{\sqrt{2\pi}} \tag{4} \\
\int_0^1 \frac{\log(1-t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt&=\frac{2(-8+\pi+2\log 2)\Gamma \left( \frac{3}{4}\right)^2}{\sqrt{2\pi}} \tag{5}
\end{align*}
$$
Substituting the results of equations $(2),(3),(4)$ and $(5)$ in $(1)$ gives
$$Q=\frac{\Gamma\left(\frac34\right)^{-2}\pi^2\log2-\Gamma\left(\frac34\right)^2(4-2\log2)}{\sqrt{2\,\pi}}$$
A: $$Q=\frac{\Gamma\left(\frac34\right)^{-2}\pi^2\ln2-\Gamma\left(\frac34\right)^2(4-\ln4)}{\sqrt{2\,\pi}}$$
