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
$$0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x^3}}$$
 A: 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.
A: 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).
A: First of all, it's your right to choose different comparison functions, since your problem was split and reduced to the analysis of two "subproblems". Moreover, you are also forced to consider different comparison functions, because you're studying convergence on different intervals!
As you know, $\displaystyle \int x^{-\alpha}$ is convergent near 0 iff $\alpha < 1$, while it is convergent near $+\infty$ iff $\alpha>1$. Thus you see that
$$ \dfrac{1}{\sqrt{x^3+x}}\leq \dfrac{1}{\sqrt{x^3}}=x^{-\frac32}$$ is unuseful near 0, because $\displaystyle \int_0^1 x^{-\frac32}$ goes to infinity. By the comparison test, you can't say anything about the finiteness of something which is just $< +\infty$.
