evaluate $\int_{0}^{1}\frac{\text{Li}_2(\frac{x^2-1}{4})}{1-x^2}dx$ I came across this integral:
$$\int_{0}^{1}\frac{\text{Li}_2(\frac{x^2-1}{4})}{1-x^2}dx=\frac{1}{2}\int_{-1}^{1}\frac{\text{Li}_2(\frac{x^2-1}{4})}{1-x^2}dx$$
One way to evaluate is to start with the subsitituion $x=2(v-0.5)$ to get:
$$\int_{0}^{1}\frac{\text{Li}_2((v-1)v)}{v(1-v)}dv$$
We can use the series expansion of $\text{Li}_2((v-1)v)$ & beta function to get the A.A Markov series which equals $-\frac{4}{5}\zeta(3)$ which is:
$$\sum_{n=1}^{\infty} \frac{2(-1)^{n}}{n^3\begin{pmatrix}2n\\ n\end{pmatrix}}$$
We can then prove that this sum equals $-\frac{4}{5}\zeta(3)$ by elementary means.
My question: is there anther approach to evaluate this integral besides using A.A Markov series? Like an identity of the dilogarithm function or integral manipulations?
 A: Substitute $x=\sin y$ and we have
\begin{align}\int_0^{\frac{\pi}{2}}\hbox{Li}_2\left(-\frac{\cos^2 y}{4}\right)\frac{dy}{\cos y}&=-\int_0^{\frac{\pi}{2}}\left(-\frac{\cos^2y}{4}\right)\int_0^1\frac{\log(t)}{1+\frac{t\cos^2 y}{4}}\frac{dtdy}{\cos y}\\
&=\int_0^1\log(t)\int_0^{\frac{\pi}{2}}\frac{\cos y}{4+t\cos^2 y}dydt\\&=\int_0^1\log(t)\int_0^{\frac{\pi}{2}}\frac{\cos y}{4+t\left(1-\sin ^2 y\right)}dydt\\&=\int_0^1\log(t)\tanh^{-1}\left(\sqrt{\frac{t}{t+4}}\right)\frac{dt}{\sqrt{t^2+4t}}\\&=4\int_0^1\log(t)\tanh^{-1}\left(\frac{t}{\sqrt{t^2+4}}\right)\frac{dt}{\sqrt{t^2+4}}\\&\overset {\text{IBP}}=-2\int_0^1\left(\tanh^{-1}\left(\frac{t}{\sqrt {t^2+4}}\right)\right)^2\frac{dt}{t}\\&=-2\int_0^1\frac{\hbox{arcsinh}^2\left(\frac{t}{2}\right)}{t}dt\\&=-2\int_0^{\frac{1}{2}}\frac{\hbox{arcsinh}^2(y)}{y}dy\end{align}
The last integral is equal to series you have mentioned, however, for different approaches see here or here which is equal to $\frac{\zeta(3)}{10}$ and hence answer of original problem is $-\frac{\zeta(3)}{5}$.
We used the following identities
$$\hbox{Li}_2(z)=-z\int_0^1\frac{\log(t)}{1-zt}dt \\
\hbox{arcsinh}(x)=\hbox{arctanh}\left(\frac{x}{\sqrt {1+x^2}}\right)$$
