Find power series representation of $ x/(x^{2}+9)^{2}$

Question

I'm not sure how to do it since the entire bottom term is squared. Is there a geometric series I should use? Or differentiation?

Answer 1

$$\int\frac x{(x^2+9)^2}dx=-\frac1{2(x^2+9)}=-\frac1{18}\cdot\frac1{\left(1+\dfrac{x^2}9\right)}$$

$$=-\frac1{18}\left(1+\frac{x^2}9\right)^{-1}$$

Answer 2

Hint

If you just look for a series representation, just write $$\frac{x}{(x^{2}+9)^{2}}=\frac{x}{81} \frac{1}{\Big(1+\frac{x^2}{9}\Big)^2}=\frac{x}{81} \frac{1}{(1+y)^2}$$ where $y=\frac{x^2}{9}$. Now, remember that $$\frac{1}{(1+y)^2}=-\frac{d}{dy}\Big(\frac{1}{1+y}\Big)$$ So, expand the last term as usual in terms of $y$, differentiate with respect to $y$, replace $y$ by $\frac{x^2}{9}$ and multiply by $\frac{x}{81}$.

I am sure that you can take from here.