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? 

Your function $\frac{\sqrt[3]x}{1+x}\sim \frac{\sqrt[3]x}{x}=\frac{1}{x^{2/3}}$ as $x\rightarrow +\infty$. We now that the integral $\int_{c}^{\infty}\frac{1}{x^{\alpha}}$ converges for $a>1$ (with $c>0$). Since $2/3 < 1$, your integral diverges
