Help with $\int_{0}^{\infty}\frac{1}{\sqrt{x^{5}+1}}dx$ $$\int_{0}^{\infty}\frac{1}{\sqrt{x^{5}+1}}dx$$
So far,
$$\int\frac{1}{\sqrt{x^{5}+1}}dx=\int\frac{1}{\sqrt{x^{4}\left(x+\frac{4}{x^{4}}\right)}}dx=\int\frac{1}{x^{2}\sqrt{x+\frac{4}{x^{4}}}}dx$$
but from there I do not know how to proceed, I have tried several
changes of variable but I have not obtained anything clear. 
I only can work with real methods. Any suggestion? Thanks!
Edit: It's sufficiente to prove that it converges

I would start by breaking the integral into two pieces
\begin{align*}
\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} = \int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} + \int_{1}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}}
\end{align*}
Thus, according to the change of variables $u = 1/x$, one has that
\begin{align*}
\int_{1}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} = \int_{0}^{1}\frac{\sqrt{u}}{\sqrt{u^{5}+1}}\mathrm{d}u
\end{align*}
Therefore 
\begin{align*}
\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^{5}+1}} = \int_{0}^{1}\frac{\sqrt{x}+ 1}{\sqrt{x^{5}+1}}\mathrm{d}x
\end{align*}
Since the integrand is continuous on $[0,1]$, the integral converges.
Hopefully this helps.
