TITLE: Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: \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.