Integrate $\int_{0}^{1}\frac{\ln(2-x)}{2-x^{2}}dx$ a) Calculate the following integral:
$ \displaystyle\int_{0}^{1}\frac{\ln(2-x)}{2-x^{2}}dx $;
b) Prove that $ f(1+0)=\ln 3 $, where $ f(x)=\displaystyle\int_{x}^{x^{3}}\frac{1}{\ln t}dt $.
I tried solving them by substitution or partial integration, but none of them worked for me.
 A: 
Let $I$ denote the value of the definite integral,
$$I:=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2}}\,\mathrm{d}x\approx0.215993.$$
One could of course calculate $I$ by brute force in terms of dilogarithms and then employ polylog identities to reduce the number of independent dilogarithm terms appearing in the result as much as possible. But this strategy is like cracking peanuts with a sledgehammer. The final result is in fact elementary, and clever use of symmetry can avoid any mention of dilogs altogether:
$$\begin{align}
I
&=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2}}\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{\ln{\left(1+t\right)}}{2-\left(1-t\right)^{2}}\,\mathrm{d}t;~~~\small{\left[x=1-t\right]}\\
&=\int_{0}^{1}\frac{\ln{\left(1+t\right)}}{1+2t-t^{2}}\,\mathrm{d}t\\
&=\int_{1}^{0}\frac{\left(1+u\right)^{2}\ln{\left(\frac{2}{1+u}\right)}}{2\left(1+2u-u^{2}\right)}\cdot\frac{\left(-2\right)}{\left(1+u\right)^{2}}\,\mathrm{d}u;~~~\small{\left[t=\frac{1-u}{1+u}\right]}\\
&=\int_{0}^{1}\frac{\ln{\left(\frac{2}{1+u}\right)}}{1+2u-u^{2}}\,\mathrm{d}u\\
&=\int_{0}^{1}\frac{\ln{\left(2\right)}}{1+2u-u^{2}}\,\mathrm{d}u-\int_{0}^{1}\frac{\ln{\left(1+u\right)}}{1+2u-u^{2}}\,\mathrm{d}u\\
&=\ln{\left(2\right)}\int_{0}^{1}\frac{\mathrm{d}u}{1+2u-u^{2}}-I,\\
\end{align}$$
and thus,
$$\begin{align}
I
&=\frac12\ln{\left(2\right)}\int_{0}^{1}\frac{\mathrm{d}u}{1+2u-u^{2}}\\
&=\ln{\left(\sqrt{2}\right)}\int_{0}^{1}\frac{\mathrm{d}v}{2-v^{2}};~~~\small{\left[u=1-v\right]}\\
&=\frac{\ln{\left(\sqrt{2}\right)}}{\sqrt{2}}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\mathrm{d}w}{1-w^{2}};~~~\small{\left[v=\sqrt{2}\,w\right]}\\
&=\frac{\ln{\left(\sqrt{2}\right)}}{\sqrt{2}}\tanh^{-1}{\left(\frac{1}{\sqrt{2}}\right)}.\blacksquare\\
\end{align}$$
A: Note that 
$$I = \int_{0}^{1}\frac{\ln(2-x)}{2-x^{2}}dx = \frac{1}{2\sqrt{2}} \left( \int_{0}^{1}\frac{\ln(2-x)}{(\sqrt{2}-x)}dx + \int_{0}^{1}\frac{\ln(2-x)}{(\sqrt{2}+x)} dx \right)$$
The first integral could be rewritten as 
$$I_1 = \int^{\sqrt{2}}_{\sqrt{2}-1}\frac{\ln(2-\sqrt{2})+\log(1+t/(2-\sqrt{2}))}{t}dt$$
Now using 
$$\mathrm{Li}_2(x) = - \int^x_0 \frac{\log(1-t)}{t} \,dt$$
We get 
$$I_1 = \log(2-\sqrt{2})\log(\sqrt{2}(\sqrt{2}+1)+\mathrm{Li}_2(-1/\sqrt{2}) - \mathrm{Li}_2(-1 - \sqrt{2})$$
Similarly 
$$I_2 = -\log(2) \log(\sqrt{2}-1
) + \mathrm{Li}_2(1 - 1/\sqrt{2}) - \mathrm{Li}(2 - \sqrt{2})$$
Finally we have 
$$I_1+I_2 = -\log(2) \log(\sqrt{2}-1 ) + \mathrm{Li}_2(1 - 1/\sqrt{2}) -\mathrm{Li}_2(2 - \sqrt{2})+\\\log(2-\sqrt{2})\log(\sqrt{2}(\sqrt{2}+1))+\mathrm{Li}_2(-1/\sqrt{2}) -\mathrm{Li}_2(-1 - \sqrt{2}))$$
AND 
$$I = \frac{I_1+I_2}{2\sqrt{2}}$$
