Integral $\int_{0}^{1}\frac{\log x}{2 - x} dx$ 
Calculate: $$\int_{0}^{1}\frac{\log x}{2 - x} dx$$

I've done a lot of research here in the community. I tried varying variants, but I did not get anything. The problem is the $2$ present in the integral.

\begin{align}J&=\int_{0}^{1}\frac{\log x}{2 - x} dx\\
\end{align}
Perform the change of variable $x=2y$,
\begin{align}J&=\int_{0}^{\frac{1}{2}}\frac{\log(2y)}{1 - y} dy\\
&=\ln 2\left[-\ln(1-x)\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y} dy\\
&=\ln ^2 2+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y} dy\\
&=\ln ^2 2+\left[-\ln(1-y)\ln y\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{\ln(1-y)}{y}\,dy\\
&=\int_{0}^{\frac{1}{2}}\frac{\ln(1-y)}{y}\,dy\\
\end{align}
Perform the change of variable $u=1-y$,
\begin{align}J&=
\int_{\frac{1}{2}}^1\frac{\ln u}{1-u}\,du\\
&=\int_{0}^1\frac{\ln u}{1-u}\,du-\int_0^{\frac{1}{2}}\frac{\ln u}{1-u}\,du\\
&=\int_{0}^1\frac{\ln u}{1-u}\,du+\ln^2 2-J\\
\end{align}
Therefore,
\begin{align}J&=\frac{1}{2}\int_{0}^1\frac{\ln u}{1-u}\,du+\frac{1}{2}\ln^2 2\\
&=\frac{1}{2}\int_0^1\left( \sum_{n=0}^\infty u^n\right)\ln u\,du+\frac{1}{2}\ln^2 2\\
&=\frac{1}{2}\sum_{n=0}^\infty\left(\int_0^1 u^n\ln u\,du\right)+\frac{1}{2}\ln^2 2\\
&=\frac{1}{2}\ln^2 2-\frac{1}{2}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\
&=\boxed{\frac{1}{2}\ln^2 2-\frac{1}{2}\zeta(2)}\\
\end{align}
