Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$ How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{2},\frac{3}{4};\frac{7}{4};-1\right)$. Can this be evaluated analytically? Any help will be appreciated.

As I said in my comment I believe by making the substitution:
$$x^2=\tan(t)$$
you get:
$$\int_0^1\frac{x^2}{\sqrt{x^4+1}}dx=\frac{1}{2}\int_0^{\pi/4}\sin(t)^{1/2}\cos(t)^{-3/2}dt=\frac{1}{4}\int_0^{1/\sqrt[4]{2}}u^{-1/4}(1-u)^{-5/4}du=B\left(2^{-1/4};\frac34,-\frac14\right)$$
