<s>[INST] Prove through induction that if $f(x) := \frac{2}{1-x^2}$ then $f^{(n)}(x) = n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$ The original formula is $f(x) := \frac{2}{1-x^2}$, for $f\colon\mathbb{R}- ({-1,1}) \to \mathbb{R}$
Through Induction prove that $f^{(n)}(x)=n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$.
So I start of with $n=0$ as the base case and see that:
$$f^{(0)}(x)=0!(\frac{1}{(1-x)^{0+1}}+\frac{(-1)^0}{(1+x)^{0+1}})$$
$$f^{(0)}(x)=(\frac{1}{(1-x)}+\frac{1}{(1+x)})$$
$$f^{(0)}(x)=\frac{1(1+x)+1(1-x)}{(1-x)(1+x)}$$
$$f^{(0)}(x)=\frac{2}{(1-x)(1+x)}$$
$$f^{(0)}(x)=\frac{2}{1-x^2}$$
Then I presumed that because it works for $n=0$ then it must also work for all $n∈\mathbb N$. Then I proceeded to state that it must also work for $n=n+1$.
$$f^{(n+1)}(x)=(n+1)!(\frac{1}{(1-x)^{n+1+1}}+\frac{(-1)^{n+1}}{(1+x)^{n+1+1}})$$
$$f^{(n+1)}(x) = f^{(n)'}(x)\\$$
$$=[(n!)((\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}}))]´$$
$$=(n!)[(\frac{1}{(1-x)^{n+1}}]´ + [\frac{(-1)^n}{(1+x)^{n+1}}]´$$
$$=n!(-(-n-1)(1-x)^{-n-2})+((-1)^n(-n-1)(1+x)^{-n-2})$$
But I'm kinda stuck on how to continue on from there ... 
 [/INST] you did all right until the last line 
$$....=(n!)\left[(\frac{1}{(1-x)^{n+1}}\right]' + \left[\frac{(-1)^n}{(1+x)^{n+1}}\right]'  =(n!)(\frac{n+1}{(1-x)^{n+2}} - \frac{(n+1)(-1)^n}{(1+x)^{n+2}} \\= (n+1)!\left(\frac{1}{(1-x)^{n+2}} + \frac{((-1)^{n+1}}{(1+x)^{n+2}}\right)$$
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