Can I estimate this integral like that? I have the following integral $ \int_{2}^{\infty} \frac{1}{\sqrt[3]{x^{3}-1}} d x $ and I should solve it without calculate it directly.
So if I find two function which is smaller and bigger than $ \int_{2}^{\infty} \frac{1}{\sqrt[3]{x^{3}-1}} d x $ and those converges, then my integral has to converge too.
So i would like to choose $ 0 \le  \frac{1}{\sqrt[3]{x^{3}-1}} d x \geq   \frac{1}{\sqrt[3]{x^{5}}} $ is this possible ?
 A: The improper integral
$$\int_2^{\infty} \frac{1}{\sqrt[3]{x^3-1}} dx$$
diverges. We find that
$$\frac{1}{\sqrt[3]{x^3-1}} > \frac{1}{x}$$
for all $x \in [2,\infty)$, and because $\int_2^{\infty}\frac{1}{x}dx$ diverges, we can conclude by the direct comparison test that the orginal improper integral is also divergent.
A: As others have pointed out this integrals is divergent. It would be
nice to know this a priori so that we have some idea on how to go
about constructing a proof of the fact. Here is a way to think about
doing so.
Scratch work/Reasoning/hand-waving/ :).
Integrals whose limits of integration involve $\pm \infty$ are called
improper integrals of type I. Whether such an integrals diverges or
converges depends on what happens to the tail of the function. In
our case we want to think about what happens to
$\frac{1}{\sqrt[3]{x^3-1}}$ when $x$ is very large. Indeed when $x$ is
very large (positive) $x^3 -1 \approx x^3$. Therefore
\begin{align}
 \frac{1}{\sqrt[3]{x^3 -1}} \approx \frac{1}{\sqrt[3]{x^3}} = \frac{1}{x}
\end{align}
for very large $x$.
Now instead of thinking about the difficult function we think of
this simplified version. What do we know about
\begin{align}
\int_2^\infty \frac{1}{x}
\end{align}
We know from our knowledge about $p$- integrals that it diverges. So
we expect our original integral to diverge as well. It is now time
to make this more precise. But now that I know what I would like
to show I can go about constructing the appropriate inequality. 
To be able to invoke the comparison theorem for improper integrals
I would like
\begin{align}
\frac{1}{x} \stackrel{?}{\leq} \frac{1}{\sqrt[3]{x^3 -1}} \;\;,
\text{for}\; x \geq 2.
\end{align}
Let's reverse engineer this and see if it holds water. Well,
\begin{align}
& \frac{1}{x} {\leq} \frac{1}{\sqrt[3]{x^3 -1}} \\
& \iff \sqrt[3]{x^3 -1} \leq x, \quad \text{cross multiplying} \\
& \iff x^3 - 1 \leq x^3, \quad \text{taking cubes of both sides} \\
& \iff x^3 \leq x^3 + 1
\end{align}
The last inequality is certainly true for $x \geq 2$. We are now in a
position to write a proper mathematical argument to show what we intended.

Proof
We have,
$$
\frac{1}{x} \leq \frac{1}{\sqrt[3]{x^3 -1}},
$$
for $x \geq 2$. We also know that $\displaystyle \int_2^\infty
                                   \frac{1}{x} \; dx$
is divergent. By the comparison theorem for improper integrals
we can now conclude that
$$
\int_2^\infty \frac{1}{\sqrt[3]{x^3 -1}}\;dx,
$$
also diverges.

The type of asymptotic reasoning we did in our scratch work is
helpful in giving us some hints as to how we should proceed with
our reasoning. i.e., Are we trying to prove convergence or
divergence? (and consequently exactly which direction should our
inequalities be etc).
To see if you understood see if you can answer the following:

Exercise
Determine if 
$$
\int_2^\infty \frac{1}{\sqrt{x^5 + x^3 + x^2 + 1}}\; dx
$$
is convergent or divergent. Justify your answer.
