Improper integral, divergence/convergence

[usn7564]

Homework Statement

Evaluate if the integral diverges or converges using the comparison theorem.

$$\int^ \infty_2 \frac{dx}{\sqrt{x^3+1}}$$

Having trouble with this question, the exercises I have managed I generally guessed if it was convergent/divergent, and then found a smaller of bigger function at the interval and went by that. Here I assumed it was convergent and used $$\frac{1}{\sqrt{x^2+1}}$$, though that was divergent which says absolutely nothing. Ran into the same issue with other attempts.

Is my reasoning the issue or have I just not found the correct manipulation? Not really approaching with anything more involved than "Find something bigger and hope it converges".
Any help would be appreciated, thanks.

 

[HallsofIvy]

It seems to me that a simpler comparison is $x^3< x^3+ 1$ so that $\frac{1}{\sqrt{x^3+ 1}}< \frac{1}{\sqrt{x^3}}= x^{-3/2}$.

 

[usn7564]

I feel slightly moronic now, was too focused on manipulating the x term for whatever reason.

Thanks, glad I didn't get it all backwards at least.