Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$ 
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$

I know that I should use Taylor's theorem and create power series. However I don't have idea how I can find $a_{n}$ such that $f(x)=\sum_{n=1}^{+\infty} a_{n}x^{n}$. I know only that $f(x)=\frac{x}{(x-1)(x^{2}+2)}$ but don't know how I can continue it because $f(x)=\frac{A}{(x^2+2)}+\frac{B}{(x-1)}$ and I get that $A=1, B=0, A=2B$  so it is conflict.Have you some tips how I can write this function and finish this task? Updated: Thanks to @gt6989b I know that $f(x)=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot \frac{1}{1-x}=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot (1+x+x^{2}+...)$ but  I still have a problem with $\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}$
 A: $$
\begin{split}
f(x) &= \frac{x}{x^3-x^2+2x-2} 
      = \frac{x}{x^2(x-1)+2(x-1)}
      = \frac{x}{(x^2+2)(x-1)}
\end{split}
$$
and use partial fractions. Then use
$$
\frac{1}{1-u} = 1 + u + u^2 + \ldots
$$
and
$$
\frac{1}{1+u} = \frac{1}{1-(-u)} = 1 -u + u^2-u^3 \pm \ldots
$$
UPDATE
From partial fractions,
$$
\frac{Ax+B}{x^2+2} + \frac{C}{x-1}
 = \frac{(Ax+B)(x-1) + C(x^2+2)}{(x^2+2)(x-1)}
$$
expanding the numerator and equating to the desired expression you get
$$
x = (Ax+B)(x-1) + C(x^2+2)
  = x^2(A+C) + x(-A+B) -B+2C
$$
which results in the system of 3 equations and 3 unknowns
$$
\begin{cases}
A &    & + C & = 0 \\
-A& +B &     & = 1 \\
  & -B & +2C & = 0
\end{cases}
$$
Adding all three together yields $C = 1/3$ which implies $A = -1/3$ and $B = 2/3$.
Can you now finish this problem?
UPDATE 2
Another hint:
$$
\frac{Ax+B}{x^2+2}
 = \left[\frac{Ax}{2} + \frac{B}{2}\right]
   \frac{1}{1 + (x/\sqrt{2})^2}
$$
and now the fraction expands as $(1+u)^{-1}$ and then expand the bracket by multiplying by the resulting Taylor series, getting 2 different series, i.e.
$$
(ax+b) \sum_{k=0}^\infty a_k x^k
 = ax \sum_{k=0}^\infty a_k x^k + b \sum_{k=0}^\infty a_k x^k
 = \sum_{k=0}^\infty (a a_k) x^{k+1} + \sum_{k=0}^\infty (b a_k) x^k
$$
and now change the index on the left summation and combine to get $\sum_{j=0}^\infty c_j x^j$ and you need $c_{22}$...
A: By way of a contrast, here's an intuitve method that lets you see the bigger picture rather than getting bogged down in technical details,
$$f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$$
$$=\frac{1}{1-x}\times \frac{x}{2+x^2}$$
$$=\frac{1}{1-x}\times \frac{x}{2(1+(\frac{x}{\sqrt{2}})^2}$$
$$=\frac{1}{1-x}\times \frac{1}{\sqrt{2}}\times\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}$$
Now, standard result,
$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}+\dots$$
$$\frac{x}{1+x^2}=x-x^3+x^5-x^7+x^9-x^{11}+\dots$$
$$\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}=\frac{x}{\sqrt{2}}-\big(\frac{x}{\sqrt{2}}\big)^3+\big(\frac{x}{\sqrt{2}}\big)^5-\big(\frac{x}{\sqrt{2}}\big)^7+\big(\frac{x}{\sqrt{2}}\big)^9-\big(\frac{x}{\sqrt{2}}\big)^{11}+\dots$$
$$\frac{1}{\sqrt{2}}\times\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}=\frac{x}{2}-\frac{x^3}{4}+\frac{x^5}{8}-\frac{x^7}{16}+\dots$$
Another standard result; multiplying by $\frac{1}{1-x}$ forms the partial sums,
$$\frac{x}{x^{3}-x^{2}+2x-2}=\frac{x}{2}+\frac{x^3}{4}+\frac{3x^5}{8}+\frac{5x^7}{16}+\frac{11x^9}{32}+\dots$$
and I'll leave it to the reader to fight through to the end of the question.
