Find the Maclaurin Series for $\frac{x}{x^2 + 1}$ How to find the Maclaurin Series for $$\frac{x}{x^2 + 1}?$$

I know that the Maclaurin Series is given by:
$$f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dots + \dfrac{f^{(k)}(0)}{k!}x^k.$$
So from repeatedly deriving $f(x)$ until there was a pattern. I have found that the Maclaurin Series for $\frac{1}{1+x}$ is given as $1-x + x^2 - x^3 + x^4 - x^5 + \dots$
In sigma notation:
$$\sum_{n=0}^{\infty} (-1)^n \cdot x^n.$$

Now I need to get the result for the new function $\frac{x}{x^2 + 1}$. Because $x^2+1$ is irreducible is there any way to do this without getting into imaginary numbers? Formally, the question says "Using your series from $\frac{1}{1+x}$, find the Maclaurin series for $\frac{x}{x^2+1}$" Perhaps I have to multiply through by $x^2$? 

$${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-1)^nx^{n}$$
let $x\rightarrow x^2$
$${\frac {1}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n}$$
then multiply by $x$
$${\frac {x}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n+1}$$
