Prove $\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$ I managed here to prove $$\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$$
but what I did was converting the LHS integral to a series then converting the series to the RHS integral. Is it possible to relate the two integrals without going through the series?
 A: First consider the following definition:
$$\operatorname{Li}_2\left(x\right)=-\int _0^x\frac{\ln \left(1-t\right)}{t}\:dt.$$
This means that:
$$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:dx=-\int _0^1\frac{1}{\sqrt{1-x^2}}\underbrace{\int _0^{-x^2}\frac{\ln \left(1-t\right)}{t}\:dt}_{y=-\frac{t}{x^2}}\:dx$$
$$=-\int _0^1\frac{1}{y}\int _0^1\frac{\ln \left(1+yx^2\right)}{\sqrt{1-x^2}}\:dx\:dy.$$
Now if we use differentiation under the integral sign for that integral we obtain:
$$=-\int _0^1\frac{1}{y}\left(\pi \ln \left(\frac{1+\sqrt{1+y}}{2}\right)\right)\:dy.$$
Therefore:
$$\boxed{\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:dx=\pi \int _0^1\frac{\ln \left(\frac{2}{1+\sqrt{1+y}}\right)}{y}\:dy.}$$
A: From the linked post we know that:
$$I=\int_0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}} dx=\pi\int_0^1\frac{\ln y}{2y}\left(1-\frac{1}{\sqrt{1+y}}\right)dy$$
So we just need to integrate by parts. We have:
$$\int \frac{1}{2y}\left(1-\frac{1}{\sqrt{1+y}}\right)dy\overset{\sqrt{1+y}= u}=\int\frac{du}{1+u}=\ln(1+\sqrt{1+y})+C$$
However, if we would use the above when integrating by parts we would run into divergence issues as
$\lim\limits_{y\to 0} \, \ln y\,  \ln(1+\sqrt{1+y})$ is not nice.
Therefore we will first substract $\ln 2$ from that and get $\lim\limits_{y\to 0} \, \ln y\, (\ln(1+\sqrt{1+y})-\ln 2)=0$.
$$\Rightarrow I=\pi\int_0^1 \ln y\,(\ln(1+\sqrt{1+y})-\ln 2)' dy\overset{IBP}=\pi\int_0^1\frac{\ln 2-\ln(1+\sqrt{1+y})}{y}dy$$
