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?

i think you can do a taylor series about a general center $x = a.$ of course there is more work to do. i am going to try to find a series about $x = a.$  make a change of variable $x = a + h,$ then we have 
$$\frac{x}{x^2 - 4x + 5} = \frac{a + h}{(a+h)^2 - 4(a+h) + 5} = \frac{a + h}{\left(a^2 - 4a + 5 + 2h(a-2)+h^2\right)}=u_0+u_1h+\cdots+u_nh^n+\cdots$$
let $A = a^2 - 4a + 5, B = 2(a-2),$
then we have  $$\left( u_0+u_1h+u_2h^2+\cdots+u_nh^n+\cdots \right)\left(A + Bh +h^2\right) = a + h$$ equating the coefficients of $h$ gives :
$$ Au_0 = a,\,  Au_1+Bu_0 = 1,\, Au_2+Bu_1+u_0 =0, \,Au_n+Bu_{n-1}+u_{n-2}=0, n = 3, 4, \cdots,  $$
we have the recurrence relation $$u_n = -\frac BA u_{n-1}- \frac1Au_{n-2}, \, u_0 =\frac aA,\, u_1 = - \frac {Ba}{A^2} + \frac 1A,\, u_2 = \frac {B^2a}{A^3} -\frac{B}{A^2} - \frac{a}{A^2}, \cdots $$ 
e.g.;  $a = 2, A = 1, B = 0$ and the simpler recurrence relation is $$u_n = -u_{n-2}, u_0 = 2, u_1 = 1, u_2 = -2, u_3 = -1, \cdots, u_{2n}=(-1)^n2, u_{2n-1} = (-1)^{n-1}$$  and the series is $$ \frac{x}{x^2 - 4x + 5} = 2 +(x-2) -2(x-2)^2 -(x-2)^3 +2(x-2)^4+\cdots$$
