Find coefficient of generating function f(x). Find coefficient of generating function.
$ f(x) = \frac{2x}{1-x^{2}} +x$ 

MY WAY OF SOLVING SIMILAR PROBLEM:
1) $ g(x) = \frac{2x}{1-x^{2}}$ 
2) partial fraction $g(x) = \frac{A}{1-x} + \frac{B}{1+x} $
3) $ g(x) =  \sum\limits_{n=0}^\infty Ax^{n} + \sum\limits_{n=0}^\infty B (-1)^nx^{n} = \sum\limits_{n=0}^\infty (A+(-1)^nB)x^{n} $ -solution
But what can I do with $f(x)$? I can't use my method because:
$f(x) = \frac{2x+x(1-x^2)}{1-x^2} $
$\frac{-x^3 +3x}{1-x^2} = \frac{A}{1-x} + \frac{B}{1+x}$
$ -x^3+3x = A(1+x) + B(1-x) $
$-x^3 = 0 \cdot x^3  $ 
$ -1 =0 $

There is no reason to add up the right hand terms, since the term $x$ is simple and convenient. The other term can be expanded using the geometric series expansion
\begin{align*}
\frac{1}{1-y}=\sum_{n=0}^\infty y^n\qquad\qquad |y|<1
\end{align*}
with $y=x^2$.

We obtain
  \begin{align*}
f(x)&=\frac{2x}{1-x^2}+x\\
&=2x\sum_{n=0}^\infty x^{2n}+x\\
&=3x+2\sum_{n=1}^\infty x^{2n+1}
\end{align*}
We conclude the coefficient $[x^n]$ of $f(x)$ is
  \begin{align*}
[x^n]f(x)=
\begin{cases}
3&n=1\\
2&n>1, odd\\
0&n\geq 0, even
\end{cases}
\end{align*}

