I have to see if $$\int\_{0}^{\infty}\frac{1}{\sqrt{x^3+x}}dx$$ is convergent or not.

I know I have two improper points so $$\int\_{0}^{\infty}\frac{1}{\sqrt{x^3+x}}dx = \int\_{0}^{1}\frac{1}{\sqrt{x^3+x}}dx + \int\_{1}^{\infty}\frac{1}{\sqrt{x^3+x}}dx$$

For the integral from 1 to 0 it's used the following comparison
$$0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x}}$$

and for the integral from 1 to infinity it's used
$$0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x^3}}$$

Why in one case is used $\sqrt{x}$ and in the other the $\sqrt{x^3}$? 

In both cases isn't this valid

$$0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x^3}}$$