<s>[INST] Taylor Series of $ \frac{1}{1-x^2} $ about x=2 I am trying to form a taylor series of the following:
$ \frac{1}{1-x^2} $ about $x=2$ 
I tried factoring the equation such that it becomes the following:
$ \frac{1}{{(1+x)}{(1-x)}} $
I tried to substitute $ x = h + 2 $ into the equation and obtained the following after using partial fractions to decompose the result:
$ \frac{1}{2(h+3)} - \frac{1}{2(h+1)} $
I do not know how to proceed from here.
I know I can just compute all the derivatives of the expression and evaluate them. But this would be non-trivial. Could someone please advise me on how I could solve this question?
 [/INST] Hint: Since $x = (x-2) + 2$, then $x^2 = [(x-2) + 2]^2 = (x-2)^2 + 4(x-2) + 4$ so
\begin{align*}
\frac{1}{1 - x^2} &= \frac{1}{1 - [(x-2)^2 + 4(x-2) + 4]} = \frac{1}{-(x-2)^2 -4(x-2)-3}\\
&= -\frac{1}{[(x-2) + 3][(x-2) + 1]} \, .
\end{align*}
Now use partial fractions to finish it off.
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