Why does $ \int _1 ^3 \frac x {(x^2-9)^{4/3}}dx$ diverge?

Question

Could someone please explain to me why the following integral is diverging? And how you would go about proving that it is. $$ \int_1^3 \frac{x}{(x^2-9)^{4/3}}dx $$

Answer 1

Hint:

Substituting $x^2-9=t \; \Rightarrow \; 2xdx=dt$ the integral becomes $$ \int_{-9}^0 \frac{1}{2}t^{-4/3} dt $$

so you can easily find the primitive and see that it diverge for $t \to 0$.