Show that :$\int^1_0 \frac{f(x^2)}{\sqrt{1-x^2}}dx =\frac{\pi^2}{16}$

we have :

$\arcsin(x\sqrt{i}) =\text{arctan}\left({\sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}}\right) + \text{i } \text{arctanh}\left({\sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}}\right)$

therfore:

$I = \text{Re}\left({\int^1_0 \frac{\arcsin(x\sqrt{i})}{\sqrt{1-x^2}} dx }\right)$

we have :(Easy proof using series expansion ) $\int^1_0 \frac{\arcsin(ax)}{\sqrt{1-x^2}} dx = \text{Li}_2(a)-\frac{1}{4}\text{Li}_2(a^2)$ Therfore: $I = \text{Re}\left({\text{Li}_2(\sqrt{i})-\frac{1}{4}\text{Li}_2(i)}\right) =\frac{\pi^2}{16}$