Coefficients for partial fraction decomposition

Question

I'm not sure if this question has been asked somewhere but I couldn't find an answer to it.

I need the coefficients in this partial fraction decomposition but in a specific way:$$\frac{1}{(x^2-b^2)^n}$$I know that it can decompose into:$$\frac{1}{(x^2-b^2)^n}=\frac{1}{(x-b)^n(x+b)^n}=\sum_{k=1}^n\frac{A_{k,n,b}}{(x-b)^k}+\frac{B_{k,n,b}}{(x+b)^k},$$where $A_{k,n,b}$ and $B_{k,n,b}$ are constants. I don't know what those constants are generally speaking but I can work them out for any given $n$ and $b$. I would like a formula if one exists. More specifically, I am only concerned with $b=i\sqrt{2}$ and the $A_{k,n,b}$ side of the decomposition but stating a generalized problem might yield some insight.

Answer 1

Partial answer:

$\frac{1}{(x^2-b^2)^n}=\frac{1}{(x-b)^n(x+b)^n}=\sum_{k=1}^n\left(\frac{a_k}{(x-b)^k}+\frac{b_k}{(x+b)^k}\right)$

To compute $a_1$, multiply both sides by $(x-b)^n$. and differentiate $(n-1)$ times, and substituting $x = b$ we get:

$(n-1)!a_1 = (-1)^{n-1}n(n+1)...(2n-1)\frac{1}{(2b)^{2n-1}}$

which can be simplified to:

$a_1 = (-1)^{n-1} n\binom{2n-1}{n}\frac{1}{(2b)^{2n-1}}$