Integral $\int_0^1 \frac{\ln (2-x)}{2-x^2} \, dx$ Here is an integral that I am trying to solve for quite some time. Find a closed form for the integral:
$$\mathcal{J}=\int_0^1 \frac{\log (2-x)}{2-x^2} \, {\rm d}x$$
Here is what I've done. 
\begin{align*}
\int_{0}^{1}\frac{\log (2-x)}{2-x^2} \, {\rm d}x &= \int_{0}^{1} \frac{\log (2-x)}{\left ( \sqrt{2}-x \right )\left ( \sqrt{2}+x \right )} \, {\rm d}x\\ 
 &= \int_{0}^{1}\left [ \frac{\log (2-x)}{2\sqrt{2}(x+\sqrt{2})} + \frac{\log (2-x)}{2\sqrt{2} \left ( \sqrt{2}-x \right )}  \right ] \, {\rm d}x \\ 
 &=\frac{1}{2\sqrt{2}} \left [ \int_{0}^{1} \frac{\log (2-x)}{\sqrt{2}+x}\, {\rm d}x + \int_{0}^{1} \frac{\log (2-x)}{\sqrt{2}-x} \, {\rm d}x \right ] 
\end{align*}
Now W|A is able to evaluate the last integrals. Indeed they boil down to:


*

*$\displaystyle \int_{0}^{1}\frac{\log(2-x)}{\sqrt{2}+x} \, {\rm d}x = {\rm Li}_2 \left ( 1-\frac{1}{\sqrt{2}} \right ) -{\rm Li}_2 \left ( 2-\sqrt{2} \right ) - \log 2 \log \left ( \sqrt{2}-1 \right )$

*$\displaystyle \int_{0}^{1}\frac{\log(2-x)}{\sqrt{2}-x} \, {\rm d}x = -{\rm Li}_2 \left ( 1+\frac{1}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 2+\sqrt{2} \right ) + i \pi \log 2 + \log 2 \log \left ( \sqrt{2}+1 \right )$


However I cannot simplify the dilogs here. Before this approach I had done the following:
\begin{align*}
\int_{0}^{1} \frac{\log(2-x)}{2-x^2} &=\frac{1}{2}\int_{0}^{1} \frac{\log(2-x)}{1- \frac{x^2}{2}} \, {\rm d}x \\ 
 &=\frac{1}{2}\int_{0}^{1} \frac{\log(2-x)}{1- \left ( \frac{x}{\sqrt{2}} \right )^2} \, {\rm d}x \\ 
 &= \frac{1}{2} \int_{0}^{1} \log(2-x) \sum_{n=0}^{\infty} \left ( \frac{x}{\sqrt{2}} \right )^{2n} \, {\rm d}x\\ 
 &=\frac{1}{2} \int_{0}^{1} \log(2-x) \sum_{n=0}^{\infty} \frac{x^{2n}}{2^n} \, {\rm d}x \\ 
 &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n} \int_{0}^{1} x^{2n} \log(2-x) \, {\rm d}x \\
 &=??
\end{align*}
In this approach I am unable to evaluate the integral $\displaystyle \int_0^1 x^{2n} \log (2-x) \, {\rm d}x$.
As a side note if we replace that $2-x$ with $1-x$ then we have that:
\begin{align*}
\int_{0}^{1}\frac{\ln (1-x)}{2-x^2} \, {\rm d}x &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^n} \int_{0}^{1}x^{2n} \ln (1-x) \, {\rm d}x \\ 
 &\overset{(*)}{=} -\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n} \frac{\mathcal{H}_{2n+1}}{2n+1}\\ 
 &=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}
\end{align*}
$(*)$ since $\displaystyle \int_{0}^{1}x^n \ln (1-x) \, {\rm d}x = -\frac{\mathcal{H}_{n+1}}{n+1}$. The series is quite easy to calculate although it shall still contain polylogs.
Questions:


*

*Can we simplify those polylogs in the first attempt I have made?

*Do you see another way of evaluating the original integral?



Here's a way to calculate the integral that circumvents the use of polylogarithms entirely.
Let $I$ denote the value of the integral,
$$I:=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2}}\,\mathrm{d}x.$$
Using a clever choice of substitution, we find
$$\begin{align}
I
&=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2}}\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{\ln{\left(\frac{2}{2-y}\right)}}{\left(2-y^{2}\right)}\,\mathrm{d}y;~~~\small{\left[x=\frac{2\left(1-y\right)}{2-y}\right]}\\
&=\int_{0}^{1}\frac{\ln{\left(2\right)}}{2-y^{2}}\,\mathrm{d}y-\int_{0}^{1}\frac{\ln{\left(2-y\right)}}{2-y^{2}}\,\mathrm{d}y\\
&=\ln{\left(2\right)}\int_{0}^{1}\frac{\mathrm{d}y}{2-y^{2}}-I.\\
\end{align}$$
Then,
$$\begin{align}
I
&=\frac12\ln{\left(2\right)}\int_{0}^{1}\frac{\mathrm{d}y}{2-y^{2}}\\
&=\frac{1}{2\sqrt{2}}\ln{\left(2\right)}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\mathrm{d}t}{1-t^{2}};~~~\small{\left[y=\sqrt{2}\,t\right]}\\
&=\frac{1}{2\sqrt{2}}\ln{\left(2\right)}\operatorname{arctanh}{\left(\frac{1}{\sqrt{2}}\right)}.\\
\end{align}$$
