Can anyone solve this integral? $\int_{-\infty}^{\infty}\sqrt{x^4+1}-x^2dx$

Question

I thought of this peculiar integral: $$\int_{-\infty}^{\infty}\sqrt{x^4+1}-x^2dx$$ and I'm wondering whether or not it has a closed-form solution. I can show that this integral converges using the comparison theorem, since $$\sqrt{x^4+1}-x^2 = \frac{1}{\sqrt{x^4+1}+x^2} < \frac{1}{1+x^2}$$ but with the $\sqrt{x^4+1}$ , we also know that the function itself has no elementary integral. Any ideas?

Answer 1

HINT: Start with @JamesArathoon's integral, obtained by IBP: $$\frac{4}{3}\int_0^\infty \frac{dx}{\sqrt{x^4+1}}$$ Then, after an appropriate substitution, apply this well known formula using the Gamma function: $$\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\mathrm{B}(m,n)$$