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? 
 A: As you said one has, for any $\epsilon>0$
$$\int_0^\epsilon x^\alpha<\infty\iff a>-1$$
and
$$\int_\epsilon^\infty x^\alpha<\infty \iff \alpha<-1 $$
But in your case near $0$ you have no problems as you add $+1$ in the denominator and so the function is bounded on $[0,\epsilon]$ and so the integral stays bounded also.
As $x+1\leq 2 x\iff\frac{1}{2x}\leq\frac{1}{1+x}$ for all $x$, we have a lower bound
$$\int_1^\infty\frac{\sqrt[3]x}{1+x}dx\geq\int_1^\infty\frac{\sqrt[3]x}{2x}dx=\frac{1}{2}\int_1^\infty x^{-2/3} dx,$$
which diverges by the first case as $-2/3\geq-1$.
As the integrand is positive on $[0,1]$, no cancellation can take place. Thus the whole integral diverges.
Actually one can also bound $1 \leq x+1\leq2$ on $[0,1]$ to obtain the estimate
$$\infty>\int_0^1\frac{\sqrt[3]x}{2}dx\leq \int_0^1\frac{\sqrt[3]x}{1+x}dx\leq  \int_0^1\frac{\sqrt[3]x}{1}dx<\infty.$$
by the first case, as  $1/3>-1$.
A: But does not this holds just for $\int_{c}^{\infty}\frac{dx}{x^{\alpha}}$ where $c>0$ and  $\alpha>1$?
Yes. Here is how you obtain it. Assume $c>0$ and $\alpha>1$. Then
$$
\int_{c}^{\infty}\frac{dx}{x^{\alpha}}=\lim_{M \to  \infty}\int_{c}^M\frac{dx}{x^{\alpha}}=\lim_{M \to  \infty}\left[ \frac{ x^{1-\alpha}}{1-\alpha}\right]_{c}^M=0-\frac{ c^{1-\alpha}}{1-\alpha}<\infty.
$$
Observe that $\int_{c}^{\infty}\frac{dx}{x^2}$ ($c>0$) is convergent.
A: 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
A: $0<\int_0^1 f(x)dx<1$ 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.
