Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.

I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.  

I' sure that I'm overseeing a tiny little missing link; dear colleagues, please give me a hint.
The given $f(x)$ is given by
\begin{align}
f(x) = \frac{2(x-1)}{x^2 - 2x + 4}
\end{align}
and can be seen as, where $a = 1 + \sqrt{3} i$ and $b = 1 - \sqrt{3} i$, such that $ab=4$,
\begin{align}
f(x) = \frac{-2 \, (1 - x)}{ab \, \left(1 - \frac{x}{a}\right) \left( 1 - \frac{x}{b}\right)}
\end{align}
for which
\begin{align}
\ln f(x) &amp;= \ln\left(- \frac{4}{ab}\right) + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\
&amp;= \pi i - \ln 2 + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\
&amp;= \pi i - \ln 2 - \sum_{n=1}^{\infty} \left( 1 - \frac{1}{a^{n}} - \frac{1}{b^{n}} \right) \, \frac{x^{n}}{n} \\
&amp;= \pi i - \ln 2 + \sum_{n=1}^{\infty} \left( \frac{a^{n} + b^{n}}{4^{n}} - 1 \right) \, \frac{x^{n}}{n} \\
&amp;= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{a^{n} + b^{n} - 4^{n} }{n} \, \left(\frac{x}{4}\right)^{n} \\
&amp;= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{ 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} }{n} \, \left(\frac{x}{2}\right)^{n}.
\end{align}
Now differentiating both sides leads to
\begin{align}
\frac{f'}{f} &amp;= \sum_{n=1}^{\infty} \left( 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n} \\
&amp;= \frac{1}{2} \, \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{n \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n-1} \\
&amp;= \sum_{n=0}^{\infty} \left( \cos\left(\frac{(n+1) \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n}.
\end{align}
Expanding the first few terms provides
\begin{align}
- \frac{f'}{f} = \frac{1}{2} + \frac{5 \, x}{4} + \frac{5 \, x^{2}}{4} + \frac{17 \, x^{3}}{16} + \frac{31 \, x^{4}}{32} + \mathcal{O}(x^{5}) 
\end{align}
Since
$$
x^2-2x+4=(x-1)^2+3=(x-2u)(x-2\bar{u}),
$$
with
$$
u=e^{i\frac\pi3},
$$
we have
\begin{eqnarray}
f(x)&amp;=&amp;\frac{2x-2}{(x-2u)(x-2\bar{u})}=\frac{1}{x-2u}+\frac{1}{x-2\bar{u}}=-\frac{\bar{u}}{2}\cdot\frac{1}{1-\frac{\bar{u}}{2}x}-\frac{u}{2}\cdot\frac{1}{1-\frac{u}{2}x}\\
&amp;=&amp;-\frac{\bar{u}}{2}\sum_{n=0}^\infty\left(\frac{\bar{u}}{2}\right)^nx^n-\frac{u}{2}\sum_{n=0}^\infty\left(\frac{u}{2}\right)^nx^n
=-\sum_{n=0}^\infty\left(\frac{\bar{u}}{2}\right)^{n+1}x^n-\sum_{n=0}^\infty\left(\frac{u}{2}\right)^{n+1}x^n\\
&amp;=&amp;-\sum_{n=0}^\infty\frac{u^{n+1}+\bar{u}^{n+1}}{2^{n+1}}x^n=-\sum_{n=0}^\infty\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)x^n.
\end{eqnarray}
Also
\begin{eqnarray}
\frac{f'(x)}{f(x)}&amp;=&amp;\left[\ln |f(x)|\right]'=[\ln2+\ln|x-1|-\ln|x^2-2x+4|]'\\
&amp;=&amp;\frac{1}{x-1}-\frac{2x-2}{x^2-2x+4}=-\frac{1}{1-x}-f(x)\\
&amp;=&amp;-\sum_{n=0}^\infty x^n-\sum_{n=0}^\infty\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)x^n\\
&amp;=&amp;-\sum_{n=0}^\infty\left[1+\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)\right]x^n.
\end{eqnarray}