An idea:
$$f(x)=\frac x{9+x^2}=\frac12\;\frac{\frac29x}{1+\frac{x^2}9}\implies$$
$$ \int f(x)\,dx=\frac12\log\left(1+\frac{x^2}9\right)=\frac12\sum_{n=1}^\infty(-1)^{n+1}\frac{x^{2n}}{3^{2n}}\;,\;\;\text{for}\;\;\frac{x^2}9<1\iff -3<x<3\implies$$
$$\frac x{9+x^2}=\left(\int f(x)\,dx\right)'=\frac12\sum_{n=1}^\infty(-1)^{n+1}\frac{2nx^{2n-1}}{9^n}\;,\;\;\text{for}\;\;-3<x<3$$
