Integral $\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx$ I have a trouble with this integral
$$I=\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx.$$
Could you suggest how to evaluate it?

Here is a proof of Cleo's answer.
Rewrite the integral as
$$
\begin{align*}
I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx 
\end{align*}
$$
The first integral can be computed by calculating a derivative of the beta function.
$$
\begin{align*}
&\;\int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&\stackrel{y=x^2}{=}\frac{1}{2}\int_0^1 \frac{\log(1-y)}{y^{\frac{3}{4}}\sqrt{1-y}}dy \\
&= \frac{1}{2}\frac{\partial}{\partial \alpha}\left\{B\left(\frac{1}{4},\alpha \right)\right\}_{\alpha=\frac{1}{2}}\\
&= \frac{\sqrt{\pi}\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4} \right)}\left\{ \psi_0\left(\frac{1}{2} \right)-\psi_0\left(\frac{3}{4} \right)\right\} \\
&= \frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(2\log 2-\pi \right)
\end{align*}
$$
The other integral can be evaluated by using equation $(3.22)$ of this paper.
$$
\begin{align*}
&\;\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx  \\
&\stackrel{x=\sin^2 t}{=}2\int_0^{\frac{\pi}{2}}\frac{\log(1+\sin^2 t)}{\sqrt{1+\sin^2 t}}dt\\
&= \log(2) K(\sqrt{-1}) \\
&= \log(2)\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}
\end{align*}
$$
where $K(k)$ is the complete elliptic integral of the first kind. After combining everything, one gets
$$I=\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(\log 2-\pi \right)\approx -3.20998$$
