$\int_0^\infty \left(\sqrt[4]{1+x^4}-x\right)\,\mathrm{d}x$ related to Beta function I have to solve this integral
$$
\int_0^\infty \left(\sqrt[4]{1+x^4}-x\right)\,\mathrm{d}x
$$
that should be related to Beta Function $\mathrm{B}(x,y)$ but I have no more ideas to get there… The result is, in fact, $\frac{\sqrt\pi\Gamma(5/4)}{2\Gamma(3/4)}$. How would you solve it?
 A: Enforce the substitution $x\mapsto \sqrt{\tan(x)}$ to arrive at 
$$\begin{align}
\int_0^\infty \left((1+x^4)^{1/4}-x\right)\,dx&=\frac12\int_0^{\pi/2}\left(\sqrt{\sec(x)}-\sqrt{\tan(x)}\right)\,\frac{\sec^2(x)}{\sqrt{\tan(x)}}\,dx\\\\
&=\frac12 \int_0^{\pi/2}\frac1{\cos^2(x)}\left(\frac{1}{\sqrt{\sin(x)}}-1\right)\,dx\tag1\\\\
&\overbrace{=}^{\text{IBP}}\frac14\int_0^{\pi/2}\sin^{-1/2}(x)\,dx\tag2\\\\
&=\frac18B\left(\frac12,\frac14\right)\tag3\\\\
&=\frac18\frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)}\tag4\\\\
&=\frac18\frac{\sqrt{\pi}4\Gamma(5/4)}{\Gamma(3/4)}\tag5\\\\
&=\frac{\sqrt{\pi}\Gamma(5/4)}{2\Gamma(3/4)}
\end{align}$$
as was to be shown!

NOTES:
In going from $(1)$ to $(2)$ we integrated $(2)$ by parts with $u=\left(\sin^{-1/2}(x)-1\right)$ and $v=\tan(x)$.
In going from $(2)$ to $(3)$, we used the representation of the Beta function $B(x,y)=2\int_0^{\pi/2}\cos^{2x-1}(t)\sin^{2y-1}(t)\,dt$, with $x=1/2$ and $y=1/4$.
In arriving at $(4)$, we used the relationship between the Beta and Gamma functions $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.
In going from $(4)$ to $(5)$ we used the functional relationship of the Gamma function $\Gamma(x+1)=x\Gamma(x)$ with $x=1/4$.

A: The integral $$\int_0^\infty\big((1+x^{1/\alpha})^\alpha-x\big)\,dx=\int_0^\infty\big((1+y)^\alpha-y^\alpha\big)\alpha y^{\alpha-1}\,dy=:\color{blue}{\alpha f(\alpha)}$$ is convergent for $0<\alpha<1/2$ (or even $0<\Re\alpha<1/2$) [we need the case $\alpha=1/4$]. Now $$f(\alpha)=\alpha\int_0^\infty y^{\alpha-1}\int_0^1(x+y)^{\alpha-1}\,dx\,dy,$$ and we change the order of integrations. The inner integral then becomes $$\int_0^\infty\big(y(x+y)\big)^{\alpha-1}\,dy\underset{y=xt}{=}x^{2\alpha-1}\int_0^\infty\frac{t^{\alpha-1}\,dt}{(1+t)^{1-\alpha}}=x^{2\alpha-1}\mathrm{B}(\alpha,1-2\alpha),$$ hence $\color{blue}{f(\alpha)}=\alpha\mathrm{B}(\alpha,1-2\alpha)\int_0^1 x^{2\alpha-1}\,dx\color{blue}{=\mathrm{B}(\alpha,1-2\alpha)/2}$.
