Expressing the $n-$th derivative of $f(x)=\frac{3x^2-6x+5}{x^3-5x^2+9x-5}$ In an attempt to express the $n-$th derivative of the rational function $f(x)=\frac{3x^2-6x+5}{x^3-5x^2+9x-5}$ ,
I split it into $\left( \frac{1+2i}{(x-2-i)} + \frac{1-2i}{(x-2+i)} + \frac{1}{(x-1)} \right)$ ,
then using $y= \frac{1}{ax+b} \Rightarrow y_{n}=\frac{(-1)^n n! a^n}{(ax+b)^{n+1}}$ I ended up with:
$$(-1)^n n! [(1+2i)(x-2-i)^{-n-1}+(1-2i)(x-2+i)^{-n-1} +(x-1)^{-n-1}]$$
Now I don’t know how to get back to $\mathbb{\mathbb{R}}$ from $\mathbb{\mathbb{C}}$.
What I'd like to get is a form that doesn't make use of trigonometric functions but uses the binomial theorem.
Can anyone help me achieve this? Thank you so much in advance.

From your well-asked question, I assume that you are only interested in a trick for the part with complex coefficients
$$A(x)=:(1+2i)(x-2-i)^{-n-1}+(1-2i)(x-2+i)^{-n-1}$$ or equivalently
$$A(x)=\frac{1+2i}{(x-2-i)^{n+1}}+\frac{1-2i}{(x-2+i)^{n+1}}$$
We will profit from the fact that the terms are conjugate:
$$\begin{aligned}A(x)&=\frac{(1+2i)\big((x-2)+i\big)^{n+1}+(1-2i)\big((x-2)-i\big)^{n+1}}{\big((x-2)^2+1\big)^{n+1}}\end{aligned}$$
Let us split the numerator into
$$\left[\big((x-2)+i\big)^{n+1}+\big((x-2)-i\big)^{n+1}\right]+2i\left[\big((x-2)+i\big)^{n+1}-\big((x-2)-i\big)^{n+1}\right]$$
and use the binomial theorem
$$\left[\sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot i^k + \sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot (-i)^k \right]\\+2i\cdot \left[\sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot i^k - \sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot (-i)^k\right]$$
In the first part all terms with odd $k$ fall out, in the second part with even $k.$ We obtain (please check it)
$$A(x)=\frac{2\cdot \sum\limits_{k=0}^{\lfloor\frac{n+1}{2}\rfloor}\binom{n+1}{2k}(x-2)^{n+1-2k}(-1)^k+4\cdot \sum\limits_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n+1}{2k+1}(x-2)^{n-2k}(-1)^{k+1}}{\big((x-2)^2+1\big)^{n+1}}$$
