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?

Notice that
$$\frac{1}{1 - x^2} = 1 + x^2 + x^4 + x^6 + \cdots = \sum_{k = 0}^\infty x^{2k}$$
provided $|x| < 1$. In general, however,
$$\frac{1}{1 - x} = \frac{1}{1 - a} + \frac{x - a}{(1 - a)^2} + \frac{(x - a)^2}{(1 - a)^3} + \cdots \quad \text{if} \quad |x - a| < 1.$$
Now, simply replace $x$ with $x^2$ and set $a = 2$.
The equality above follows from the formula
$$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$$
