Determine the convergenece/divergence of an improper integtal

Question

The improper integral is as follows: $$\int_{3}^{\infty}\frac{dx}{\sqrt{x^3-9} }$$

Please help determine whether divergent or convergent , I couldn't solve it.

Answer 1

There is only one potential problem:

Around $\infty$: $$ \frac 1{\sqrt{x^3-9}}\sim \frac 1{x^{3/2}} $$ and $$\int_3^\infty \frac {dx}{x^{3/2}}=\lim_{A\to\infty} [-2x^{-1/2}]_3^A<\infty $$

Now, $$ \frac 1{\sqrt{x^3-9}}\sim \frac 1{x^{3/2}}$$ and those are positive so$$ \exists A>0 \ \ x>A\Rightarrow \frac 1{\sqrt{x^3-9}}\le \frac 2{x^{3/2}} $$ $$ \int_3^\infty \frac {dx}{\sqrt{x^2-9}} \le \int_3^A \frac {dx}{\sqrt{x^2-9}}+ \int_3^\infty \frac {dx}{x^{3/2}}< \infty $$