Showing $\int_0^\infty\frac{Li_2\frac{(1-x^4)^2}{(1+x^4)^2}}{1+x^4}dx=\sqrt2\,\Re\left(\int_0^1\frac{Li_2\frac{(1+x^4)^2}{(1-x^4)^2}}{1+x^2}dx\right)$ A beautiful equality:
$$\int_{0}^{\infty }\frac{Li_{2}\left(\frac{(1-x^{4})^{2}}{(1+x^{4})^{2}}\right)}{1+x^{4}}dx=\sqrt{2} \Re \left(\int_{0}^{1 }\frac{Li_{2}\left(\frac{(1+x^{4})^{2}}{(1-x^{4})^{2}}\right)}{1+x^{2}}dx\right)$$
This question was proposed by Sujeethan Balendran in RMM(Romanian Mathematical Magazine)
I did by using complex number but i didn't by
Real method by using
$$Li _{2}(a)=-a\int_{0}^{1}\frac{\ln x}{1-ax}dx$$
I believe some of you know some nice proofs of this, can you please share it with us?
 A: $$\int_{0}^{\infty }\frac{Li_{2}(\frac{(1-x^{4})^{2}}{(1+x^{4})^{2}})}{1+x^{4}}dx $$
$$=\frac{1}{2}\int_{0}^{\infty }\frac{Li_{2}(\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}})}{\sqrt{x}(1+x^{2})}dx$$
$$substitute\quad x=tan(\frac{x}{2})$$
$$=\frac{1}{4}\int_{0}^{\pi}Li_{2}(cos^{2}(x))(\sqrt{tan(\frac{x}{2})})dx
=\frac{1}{8}\int_{0}^{\pi}Li_{2}(cos^{2}(x))(\sqrt{tan(\frac{x}{2})}+\sqrt{cot(\frac{x}{2})})dx$$
$$=\frac{-2\sqrt{2}}{8}\Re (\int_{0}^{\pi}\frac{iLi_2(cos^{2}(x))e^{ix}}{\sqrt{1-e^{2ix}}})dx$$
$$=\frac{-2\sqrt{2}}{8}\Re (\oint_  {\left | z \right |=1}\frac{Li_2(\frac{(z+\frac{1}{z})^{2}}{4})}{\sqrt{1-z^{2}}})dz\\=\frac{1}{2\sqrt{2}}\Re \int_{-1}^{1}\frac{Li_2((\frac{x^{2}+1}{2x})^{2})}{\sqrt{1-x^{2}}}dx$$
$$=\frac{1}{2\sqrt{2}}\Re \int_{-1}^{1}\frac{Li_2((\frac{x^{2}+1}{2x})^{2})}{\sqrt{1-x^{2}}}dx$$=$$\frac{1}{\sqrt{2}}\Re \int_{0}^{1}\frac{Li_2((\frac{x^{2}+1}{2x})^{2})}{\sqrt{1-x^{2}}}dx$$
$$substitute \quad x=\frac{1-x}{1+x}\\=\frac{1}{\sqrt{2}}\Re \int_{0}^{1}\frac{Li_2((\frac{x^{2}+1}{1-x^{2}})^{2})}{(1+x)\sqrt{x}}dx$$$$
so \quad =\sqrt{2}\Re (\int_{0}^{1}\frac{Li_2((\frac{x^{4}+1}{1-x^{4}})^{2})}{(1+x^{2})}dx)$$
