@integralsbot \int_0^\infty \left(\sqrt{1+x^4}-x^2\right) \,dx=\frac{\Gamma^2\left( \frac14\right)}{6 \sqrt{\pi}}

Today we will show the following result that appears in this post from @integralsbot

$\int_0^\infty \left(\sqrt{1+x^4}-x^2\right) \,dx=\frac{\Gamma^2\left( \frac14\right)}{6 \sqrt{\pi}}$

Let

$\sqrt{1+x^4}-x^2&=\sqrt{u}$

Then:

$\begin{aligned} \sqrt{1+x^4}-x^2&=\sqrt{u}\\ \sqrt{1+x^4}&=x^2+\sqrt{u}\\ 1+x^4&=x^4+2x^2\sqrt{u}+u\\ 1-u&=2x^2\sqrt{u}\\ x^2&=\frac{1-u}{2\sqrt{u}}\\ x&=\frac{1}{\sqrt{2}}(1-u)^{1/2}u^{-1/4} \end{aligned}$

And

$dx=\frac{1}{\sqrt{2}}\left(\frac12(1-u)^{-1/2}x^{-1/4}+\frac14(1-u)^{1/2}x^{-5/4} \right)(-du)$

We then get:

$\begin{aligned} \int_0^\infty \left(\sqrt{1+x^4}-x^2\right) \,dx&=\frac{1}{2\sqrt{2}}\int_0^1x^{1/4}(1-x)^{-1/2}\,dx+\frac{1}{4\sqrt{2}}\int_0^1x^{-3/4}(1-x)^{1/2}\,dx\\ &=\frac{1}{2\sqrt{2}}\operatorname{B}\left(\frac54,\frac12 \right)+\frac{1}{4\sqrt{2}}\operatorname{B} \left(\frac14 ,\frac32 \right)\\ &=\frac{1}{2\sqrt{2}}\frac{\Gamma\left(\frac54\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac74\right)}+\frac{1}{4\sqrt{2}}\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac32\right)}{\Gamma\left(\frac74\right)}\\ &=\frac{1}{2\sqrt{2}}\frac{\frac14\Gamma\left(\frac14\right)\sqrt{\pi}}{\frac34\Gamma\left(\frac34\right)}+\frac{1}{4\sqrt{2}}\frac{\Gamma\left(\frac14\right)\frac12\Gamma\left(\frac12\right)}{\frac34\Gamma\left(\frac34\right)}\\ &=\frac{\sqrt{\pi}}{6\sqrt{2}}\frac{\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}+\frac{\sqrt{\pi}}{6\sqrt{2}}\frac{\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}\\ &=\frac{2\sqrt{\pi}}{6\sqrt{2}}\frac{\Gamma\left(\frac14\right)}{\frac{\sqrt{2}\pi}{\Gamma\left(\frac14\right)}}\\ &=\frac{\Gamma^2\left( \frac14\right)}{6 \sqrt{\pi}} \qquad \blacksquare \end{aligned}$