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.
 A: An alternative approach. By termwise integration of a Maclaurin series 
$$ I=\int_{0}^{1}\frac{x^2}{\sqrt{1+x^4}}\,dx = \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{(-1)^n}{4n+3} $$
and due to the fact that $P_{2n}(0)=\left[\frac{1}{4^n}\binom{2n}{n}\right](-1)^n$, the RHS of the previous line is the value at $x=\frac{1}{2}$ of the following function:
$$ f(x) = \sum_{n\geq 0}\frac{1}{2n+3} P_{n}(2x-1). $$
Now we may exploit three well known Fourier-Legendre expansions:
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{2}{(1-2n)(2n+3)}P_n(2x-1), $$
$$ K(x) = \sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1), $$
$$ E(x) = \sum_{n\geq 0}\frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$
(the argument $x$ stands for the elliptic modulus, like in Mathematica's notation) and deduce from partial fraction decomposition that 
$$ I = \left.\frac{1}{2}K(x)-E(x)+\sqrt{1-x}\right|_{x=1/2}. $$
$K\left(\frac{1}{2}\right)$ and $E\left(\frac{1}{2}\right)$ are well known to be related to $\Gamma\left(\frac{1}{4}\right)^2$, and related to each other via Legendre's identity.
A: \begin{align}
\int\limits_{0}^{1} \dfrac{x^{2}}{\sqrt{1+x^{4}}}\,\mathrm{d}x &= \dfrac{1}{4}\int\limits_{0}^{1} \dfrac{1}{x^{1/4}\sqrt{1+x}}\,\mathrm{d}x \\
&= \dfrac{1}{4}\int\limits_{1}^{2} \dfrac{1}{\left(x-1\right)^{1/4}\sqrt{x}}\,\mathrm{d}x \\
&= \dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{1 -\color{red}{x} + \color{red}{x}}{\left(1-x\right)^{1/4}x^{5/4}}\,\mathrm{d}x \label{subst to 1/x} \tag{1} \\
&= \color{blue}{\dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{\left(1-x\right)^{3/4}}{x^{5/4}}\,\mathrm{d}x} + \dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x \\
&\overset{\mathrm{IBP}}{=} \color{blue}{\left.-\dfrac{\left(1-x\right)^{3/4}}{x^{1/4}}\right\vert_{1/2}^{1}-\dfrac{3}{4}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x} + \dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x \\
&= \dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x \\
&= \dfrac{1}{\sqrt{2}}-\dfrac{1}{4}\operatorname{B}\left(\dfrac{3}{4},\dfrac{3}{4}\right)
\end{align}
At \eqref{subst to 1/x} substitution $x \to x^{-1}$ was made.
A: As I said in my comment I believe by making the substitution:
$$x^2=\tan(t)$$
you get:
$$\int_0^1\frac{x^2}{\sqrt{x^4+1}}dx=\frac{1}{2}\int_0^{\pi/4}\sin(t)^{1/2}\cos(t)^{-3/2}dt=\frac{1}{4}\int_0^{1/\sqrt[4]{2}}u^{-1/4}(1-u)^{-5/4}du=B\left(2^{-1/4};\frac34,-\frac14\right)$$
