What is the intergal of $\int {\dfrac{1}{(x^2+9)^2}}\,dx$ with recursive formula?

Question

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?

Answer 1

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}

Answer 2

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