Q:

Improper integral $\int _3^{\infty }\:\frac{\sqrt{x^3+2}}{\left(x^2-3x\right)^\phi}$

$$\gamma = \:\frac{\sqrt{x^3+2}}{\left(x^2-3x\right)^\phi}$$
$$\int _3^{\infty }\gamma$$
I have to find for whom the $\phi > 0$ the improper integral converges, but i'm not very familiar with this type of integral. I tried like that:  
$$\gamma \approx_{\infty }\:\frac{\sqrt{x^3}}{x^{2\phi}}\rightarrow \int _3^{\infty }\:\frac{\sqrt{x^3}}{x^{2\phi}} = \int _3^{\infty }\:\frac{1}{x^{\frac{4\phi-3}{2}}}$$
That converges for $\color{red}{\phi > 5/4}$
Then
$$\gamma \approx_3 \frac{1}{\left(x^2-3x\right)^\phi} \rightarrow \int _3^{\infty }\:\frac{1}{\left(x^2-3x\right)^\phi}$$
But i don't know how to solve it, can you give me a clue as to how to continue? Thanks

A:

Hint. As $x \to 3^+$, one has
$$
\frac{\sqrt{x^3+2}}{\left(x^2-3x\right)^\phi}= \frac{\sqrt{x^3+2}}{x^\phi(x-3)^\phi}\sim \frac{\sqrt{29}}{3^\phi}\cdot \frac1{(x-3)^\phi}
$$ this integrand is convergent near $x=3^+$ iff $\phi<1$.

