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 $\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}$

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.
