How to choose the most significant denominator

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

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

Even though the inequality is true, the fact that $\displaystyle 0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x^3}}$ won't tell you much about the integral from $0$ to $1$ since $\displaystyle \int_ 0 ^1 \frac{1}{\sqrt{x^3}}dx$ diverges, because $\displaystyle 1\leq \frac{3}{2}$.

Let $a,b,\alpha$ be real numbers such that $a<b$.

The integrals $\displaystyle \int \limits_a^{b^-} \frac{1}{(b-x)^\alpha}\mathrm dx , \int \limits_{a^+}^b \frac{1}{(x-a)^\alpha}\mathrm dx$ converge if, and only if, $\alpha <1$.

To verify this you just need to compute the antiderivatives and take the limits.

The last inequality you give is valid, but $1/\sqrt{x^3}=x^{-3/2}$ is not integrable near $x=0$.

Think of it this way: When $x$ is very small, $x^3\ll x$ (in words: $x^3$ is much smaller than $x$), so $x^3+x$ is very close to $x$ (in relative terms).

On the other hand, when $x$ is very large, $x^3\gg x$, so that $x^3+x$ is very close to $x^3$ (again, in relative terms).