The Taylor series for $1/x$ centered at $c = 1$ is:
$$1 - (x-1) + (x-1)^2 - (x-1)^3 + ...$$
When the function is shifted, to $\frac{1}{x-1}$, then we can use composition under the condition $c=2$ :
$$1 - ((x-1)-1) + ((x-1)-1)^2 - ((x-1)-1)^3 + ...$$
How can we use composition to find $\frac{1}{x-1}$ centered at any other number, like $0$?
$$f(x)=f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+...$$
You should be able to expand at whichever center you like, by setting the value of $a$, but of course the function and its derivatives of all orders have to exist at the point of $a$.