Determining power series for $\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$ I'm looking for the power series for $f(x)=\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$
My approach: the given function is a combination of two problems. first i made some transformations, so the function looks easier. 
$$\frac{3x^{2}-4x+9}{(x-1)^2(x+3)})=\frac{3x^{2}-4x+9}{x^3+x^2-5x+3}$$
Now i have two polynomials. i thought the Problem might be even easier, if thinking about the function as:
$$\frac{3x^{2}-4x+9}{x^3+x^2-5x+3)}= (3x^{2}-4x+9)\cdot \frac{1}{(x^3+x^2-5x+3)}$$
Assuming the power series of  $3x^{2}-4x+9$ is just $3x^{2}-4x+9$ itself. I hoped, i could find the power series by multiplying the series of the both easier functions.. yeah, i am stuck.
$\sum\limits_{n=0}^{\infty}a_{n}\cdot x^{n}=(3x^{2}-4x+9)\cdot \ ...?... =$ Solution

You can use the partial fraction decomposition:
$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{1+\frac{1}{3}x}
$$
and sum up the series you get, which are known.
If you do the computation, you find $A=0$, $B=2$ and $C=1$, so
$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\frac{2}{(1-x)^2}+\frac{1}{1+\frac{1}{3}x}
$$
The development of $(1-x)^{-2}$ can be deduced from the fact that
$$
\frac{1}{1-x}=\sum_{n\ge0}x^n
$$
so, by deriving, we get
$$
\frac{1}{(1-x)^2}=\sum_{n\ge1}nx^{n-1}=\sum_{n\ge0}(n+1)x^n
$$
The power series for the other term is again easy:
$$
\frac{1}{1+\frac{1}{3}x}=\sum_{n\ge0}\frac{(-1)^n}{3^n}x^n
$$
so your power series development is
$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\sum_{n\ge0}\left(2n+2+\frac{(-1)^n}{3^n}\right)x^n
$$
