<s>[INST] 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?
 [/INST] Integrate as follows
\begin{align}
\int {\dfrac{1}{(x^2+9)^2}}\,dx
&= \frac19 \int {\dfrac{1}{x^2+9}}\,dx - \frac19\int {\dfrac{x^2}{(x^2+9)^2}}\,dx\\
 &= \frac19 \int {\dfrac{1}{x^2+9}}\,dx + \frac1{18}\int x\>d({\dfrac{1}{x^2+9}})\\
 &= \frac1{18} \frac x{x^2+9} + \frac1{18}\int {\dfrac{1}{x^2+9}}\,dx\\
 &= \frac1{18} \frac x{x^2+9} + \frac1{54}\tan^{-1}\frac x3+C
\end{align}
</s>