Convergence/divergence of $\int_{0}^{\infty} \frac{\sqrt[3]x}{1+x}dx$ 
$$\int_{0}^{\infty} \frac{\sqrt[3]x}{1+x}dx$$

I have read an example on book and they did the following:
$$\frac{\sqrt[3]x}{1+x^2}<\frac{\sqrt[3]x}{x}=\frac{1}{x^{\frac{2}{3}}}$$
and we know that the integral of $\frac{1}{x^{\alpha}}$ converges for $\alpha>1$
So $$\int_{0}^{\infty} \frac{\sqrt[3]x}{1+x}dx$$ converges, but does not this holds just for $\int_{c}^{\infty}\frac{1}{x^{\alpha}}$ where $c>0$ and  $\alpha>1$?
Because $\int_{0}^{\infty}\frac{1}{x^2}$ diverges? 

$0<\int_0^1 f(x)dx<1$ and $\int_1^\infty f(x)dx>\int_1^\infty \frac{1}{2x^{\frac{2}{3}}}dx=\infty$. 
Therefore, $\int_0^\infty f(x) dx$ diverges.
