Taylor series of $x/(x^2-4x+5)$ I'm supposed to find the Taylor series of this function (I can choose to center it at any A I want):
$$f(x)= x/(x^2-4x+5)$$
When I derivate, it only gets more and more confusing. How can I make any sense out of this?

The first step is to complete the square: $x^2-4x+5 = (x-2)^2 + 1$. This suggests centering the expansion at $x-2$, since then the formula for the sum of a geometric series gives
$$
\frac{1}{1 + (x-2)^2} = \sum_{i=0}^\infty (-1)^i (x-2)^{2i}.
$$
Writing $x = (x-2) + 2$ we deduce
$$
\frac{x}{1 + (x-2)^2} = \sum_{i=0}^\infty (-1)^i [2(x-2)^{2i} + (x-2)^{2i+1}].
$$
The first few terms are
$$
\frac{x}{1 + (x-2)^2} = 2 + (x-2) - 2(x-2)^2 - (x-2)^3 + 2(x-2)^4 + (x-2)^5 - 2(x-2)^6 - (x-2)^7 + \cdots.
$$
