TITLE: Prove $\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$
QUESTION [25 upvotes]: I have in trouble for evaluating following integral
$$\int_0^{\infty}  \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$$
It seems really easy, but I don't know how to handle it at all.
(The results are well known, here I tried to evaluate it but I failed)
I tried to use the relation 
$$\sqrt{1+x^{4}}-x^{2}=\frac{1}{\sqrt{1+x^{4}}+x^{2}}$$ 
but I couldn't find the desired results.

REPLY [11 votes]: Another approach :
Set
$$
x^4=\frac1t-1\quad\color{blue}{\Rightarrow}\quad x=\left(\frac{1-t}t\right)^{\large
\frac14}\quad\color{blue}{\Rightarrow}\quad dx=-\frac14t^{-\large\frac54}\left(1-t\right)^{-\large\frac34}\ dt,
$$
then the integral turns out to be
\begin{align}
\int_0^{\infty}  \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx&=\frac14\int_0^1  \left(t^{-\large\frac74}\left(1-t\right)^{-\large\frac34}-t^{-\large\frac74}\left(1-t\right)^{-\large\frac14}\right)\ dt\\
&=\frac14\left[\text{B}\left(-\frac34,\frac14\right)-\text{B}\left(-\frac34,\frac34\right)\right]\\
&\stackrel{\color{red}{[1]}}=\frac14\left[\text{B}\left(-\frac34,\frac14\right)-\color{red}0\right]\\
&\stackrel{\color{red}{[2]}}=\frac14\cdot\frac{\Gamma\left(-\frac34\right)\Gamma\left(\frac14\right)}{\Gamma\left(-\frac12\right)}\\
&\stackrel{\color{red}{[3]}}=\color{blue}{\frac{\Gamma^2\left(\frac14\right)}{6\sqrt\pi}}.
\end{align}

Notes :
$\color{red}{[1]}\;\;\;$Use property of Beta function and  Euler's reflection formula  for the gamma function
$$\text{B}(x,y)=\text{B}(x+1,y)+\text{B}(x,y+1)$$
$\quad\;\;\;$and
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$$
$\color{red}{[2]}\;\;\;\displaystyle\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$
$\color{red}{[3]}\;\;\;\displaystyle\Gamma(n)=\frac{\Gamma(n+1)}n$ and $\displaystyle\Gamma\left(\frac12\right)=\sqrt\pi$