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

Let $x = \sqrt{\tan \theta}$. Note that, $x^4 = \tan^2 \theta$ and $dx = \dfrac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$. Like this, 
$$
I = \int_{0}^{\infty}( \sqrt{1 + x^4} - x^2)dx = \int_{0}^{\pi/2} \dfrac{(\sec \theta + \tan \theta)\sec^2\theta d\theta}{2\sqrt{\tan \theta}} = \dfrac{1}{2}\int_{0}^{\pi/2} \dfrac{\cos^{1/2}\theta (1 - \sin \theta)d\theta}{\sin^{1/2}\theta \cos^3 \theta}
$$
$$
= \dfrac{1}{2}\int_{0}^{\pi/2}\cos^{-5/2}\theta\sin^{-1/2}\theta d\theta - \dfrac{1}{2}\int_{0}^{\pi/2}\cos^{-5/2}\theta\sin^{1/2}\theta d\theta = I_1 + I_2
$$
Note that, $I_1 = \dfrac{1}{2}B(-3/4,1/4)$, because
$$
(2m-1=-5/2 \ \Rightarrow \ m = -3/4 \ \text{and} \ 2n-1 = -1/2 \ \Rightarrow \ n=1/4)
$$
But, $B(m,n) = \dfrac{1}{2}\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$. Thus, 
$$
I_1 = \dfrac{1}{4}\dfrac{\Gamma(-3/4)\Gamma(1/4)}{\Gamma(-1/2)} = \dfrac{1}{4}\cdot \dfrac{(-4)}{3}\dfrac{\Gamma(1/4)\cdot \Gamma(1/4)}{(-2)\sqrt{\pi}} = \dfrac{\Gamma(1/4)^2}{6\sqrt{\pi}}
$$
If $x \to 0$, $\Gamma(x) \to \infty$, so that $I_2 = 0$. Furthermore, if $x < 0, x \neq -1,-2,\ldots$, define $\Gamma(x) = \Gamma(x + 1)/x$.
