What is the intergal of $\int {\dfrac{1}{(x^2+9)^2}}\,dx$ with recursive formula? I used partial integration and I got this:
${\dfrac{x}{(x^2+9)^2}} + 4  $$\int {\dfrac{x^2}{(x^2+9)^3}}\,dx$
According to the recursice formula I should continue like this:
${\dfrac{x}{(x^2+9)^2}} + 4  $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}\,dx$
${\dfrac{x}{(x^2+9)^2}} + 4  $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}\,dx$ - 4$\cdot 9 $$\int {\dfrac{1}{(x^2+9)^3}}\,dx$
And then end up with something like that:
${\dfrac{1}{2 \cdot 9} \cdot \dfrac{1}{x^2+9} + \dfrac{1}{2 \cdot 27}} arctg{\dfrac{x}{3}}\,$ + C
But I'm sure that I'm doing this wrong or just I'm too confused?

You have found the integral of $(x^2+9)^{-3}$ in terms of the integral of $(x^2+9)^{-2}$.  So, instead, integrate $(x^2+9)^{-1}$ by parts
