Hard Integral $\int\limits_0^1 \frac{\ln y\operatorname{Li}_2 (-y)}{1-y^2} \, dy$

Question

How can I calculate the integral? Any ideas ?

$$\int_0^1 \frac{\ln y\operatorname{Li}_2 (-y)}{1-y^2} \, dy$$

I thought about using this formula but I can not get to something, someone can prove this? $$\int_0^1 \frac{\ln^{s-1} x\operatorname{Li}_2(x)}{1-x}dx = (-1)^{s-1}(s-1)!\frac{\zeta(s)-\zeta(2s)} 2$$

Answer 1

Hint: $$\frac{\ln{y}\operatorname{Li}_2(-y)}{1-y^2} = 1/2 *\Bigl( \frac{\ln{y}\operatorname{Li}_2(-y)}{1-y}+\frac{\ln{y}\operatorname{Li}_2(-y)}{1+y}\Bigr) $$