How can I evaluate $\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$ I've been trying to find and prove that:
$$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\pi \operatorname{Li}_2\left(\frac{1-\sqrt{2}}{2}\right)-\frac{\pi }{2}\left(\ln \left(\frac{1+\sqrt{2}}{2}\right)\right)^2$$
result obtained via WolframAlpha.
Here $\operatorname{Li}_2\left(z\right)$ denotes the dilogarithm function.
However it seems very difficult to prove it, the integral equals:
$$-\frac{\pi ^3}{24}+2\int _0^1\frac{\arcsin \left(x\right)\ln \left(1+x^2\right)}{x}\:\mathrm{d}x$$
Using the series representation of the dilogarithm function yields:
$$\sum _{n=1}^{\infty }\frac{\left(-1\right)^n}{n^2}\int _0^1\frac{x^{2n}}{\sqrt{1-x^2}}\:\mathrm{d}x$$
Perhaps using the beta function can allow us to proceed further? any kind of hint or solution is well regarded.
 A: $$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\int _0^1\frac{1}{\sqrt{1-x^2}}\left(\int_0^1\frac{x^2\ln(y)}{1+x^2 y}\mathrm{d}y\right)\mathrm{d}x$$
$$=\int _0^1\ln(y)\left(\int_0^1\frac{x^2}{(1+x^2y)\sqrt{1-x^2} }\mathrm{d}x\right)\mathrm{d}y$$
$$=\int _0^1\ln(y)\left(\frac{\pi }{2y}\left(1-\frac{1}{\sqrt{1+y}}\right)\right)\mathrm{d}y$$
$$=\int _0^1\ln(y)\left(\frac{\pi }{2y}\left(-\sum_{n=1}^\infty\frac{2n\choose n}{4^n}(-y)^n\right)\right)\mathrm{d}y$$
$$=-\frac{\pi}{2}\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n}\int_0^1 y^{n-1}\ln(y)\mathrm{d}y$$
$$=\frac{\pi}{2}\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n n^2}.\qquad (*)$$
To get this sum, first replace $x$ by $-x$ in
$$\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n}\frac{x^n}{n}=-2\ln\left(\frac{1+\sqrt{1-x}}{2}\right)$$
then divide both sides by $x$ and integrate from $0$ to $1$,
$$\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n n^2}=-2\int_0^1\frac{\ln\left(\frac{1+\sqrt{1+x}}{2}\right)}{x}\mathrm{d}x$$
$$\overset{\frac{1+\sqrt{1+x}}{2}=t}{=}2\int_1^{\frac{1+\sqrt{2}}{2}}\frac{\ln(t)}{1-t}\mathrm{d}t-2\int_1^{\frac{1+\sqrt{2}}{2}}\frac{\ln(t)}{t}\mathrm{d}t$$
$$=2\operatorname{Li}_2(1-t)-\ln^2(t)\bigg|_1^{\frac{1+\sqrt{2}}{2}}$$
$$=2 \operatorname{Li}_2\left(\frac{1-\sqrt{2}}{2}\right)-\ln^2 \left(\frac{1+\sqrt{2}}{2}\right).$$

Addendum:
You can get the summation representation in $(*)$ by expanding $\operatorname{Li}_2(-x^2)$ in series as you mentioned in your question body, then using $\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x=\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac12+n\right)}{n\Gamma(n)}=\frac{\pi}{2}\frac{{2n\choose n}}{4^n}$ which follows from using the beta function and the Legendre duplication formula:
$$\Gamma\left(\frac12+n\right)=\frac{2\sqrt{\pi}\,\Gamma(2n)}{4^n\Gamma(n)}=\sqrt{\pi}n\Gamma(n)\frac{2n\choose n}{4^n}.$$
A: Substitute $x=\sin y$, we have $$\int_0^{\frac{\pi}{2}} \operatorname{Li}_2\left(-\sin^2 y\right)dy\stackrel{\rm dilog.}{=}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}\int_0^{\frac{\pi}{2}}\sin^{2k}dy$$ Since the last integral is Well-know Wallis' Integral so we have $$\frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^24^k}{2k\choose k}$$
Since the generating function of $$\sum_{k=1}^{\infty}\frac{x^n}{k^2}{2k\choose k}=2x\left(\frac{\hbox{Li}_2\left(\frac{1}{2}-\frac{1}{2}\sqrt{1-4x}\right)}{x}-\frac{\left(\log\left(\sqrt{1-4x}+1\right)-\log(2)\right)^2}{2x}\right)$$. The proof the result can be found here. Multiply both sides by $\frac{\pi}{2}$ and set $x=-1/4$ gives the desired result.
