Integral $\int\limits_0^1\frac{-2\ln(x)(1+x^2)^2-2(1-x^4)}{(1-x^2)^3}\,\mathrm{d}x$? $$\int\limits_0^1\frac{-2\ln(x)(1+x^2)^2-2(1-x^4)}{(1-x^2)^3}\,\mathrm{d}x$$
Are there any easy methods to solve this integral? I tried to solve this by series, but I don't know how to complete. I hope the answer does not contain th $\operatorname{Li}$ function.
The last answer from WolframAlpha is $\dfrac{\pi^2}{8}$.
 A: Let $f(x)$ be your integrand.
Maple gives the antiderivative as
$$ \frac{1}{2(x+1)}+\frac{1}{2(x-1)}-\frac{\text{dilog}(x+1)}{2} -\frac{\ln  \left( x \right) \ln  \left( 
x+1 \right)}{2} -{\frac {\ln  \left( x \right) x \left( x+2 \right) 
}{2\; \left( x+1 \right) ^{2}}}-\frac{{\text {dilog}} \left( x \right)}{2} +{
\frac {\ln  \left( x \right) x \left( x-2 \right) }{2\; \left( x-1
 \right) ^{2}}}+{\frac {\ln  \left( x \right) x}{2\;(x+1)}}-{
\frac {\ln  \left( x \right) x}{2\;(x-1)}}
$$
where
$$ \text{dilog}(x) = \int_1^x \frac{\ln(t)}{1-t}\; dt $$
In fact, if $F_1(x)$ contains the terms without dilog, 
$$ F_1(x) = \frac{1}{2(x+1)}+\frac{1}{2(x-1)}-\frac{\ln  \left( x \right) \ln  \left( 
x+1 \right)}{2} -{\frac {\ln  \left( x \right) x \left( x+2 \right) 
}{2\; \left( x+1 \right) ^{2}}} +{
\frac {\ln  \left( x \right) x \left( x-2 \right) }{2\; \left( x-1
 \right) ^{2}}}+{\frac {\ln  \left( x \right) x}{2\;(x+1)}}-{
\frac {\ln  \left( x \right) x}{2\;(x-1)}}$$
you can verify that 
$$F_1'(x) - f(x) = - \frac{\ln(x)}{2(x-1)} - \frac{\ln(x+1)}{2x}$$
and $$\lim_{x \to 0} F_1(x) = \lim_{x \to 1} F_1(x) = 0$$
So your integral becomes
$$ \int_0^1 \frac{\ln(x)\; dx}{2(x-1)} + \int_0^1 \frac{\ln(x+1)\; dx}{2x} $$
Using the change of variables $u=x+1$ in the second of these and combining the two, we find your integral is 
$$ \frac{1}{2}\int_0^2 \frac{\ln(x)\; dx}{x-1} $$
which I think is fairly "well-known".
