Question

How do you write the partial fraction decomposition of the rational expression `5/(x^2-1)^2`?

Answer 1

The answer is `=5/(4(x+1)^2)+5/(4(x+1))+5/(4(x-1)^2)-5/(4(x-1))`

Explanation:

Let's factorise the denominator `(x^2-1)^2=(x+1)(x-1(x+1)(x-1)`
So `5/(x^2-1)^2=5/((x+1)^2(x-1)^2)`
`=A/(x+1)^2+B/(x+1)+C/(x-1)^2+D/(x-1)`
`5=` `A(x-1)^2+B(x+1)(x-1)^2+C(x+1)^2+D(x-1)(x+1)^2`
Let `x=1` then `5=4C` `=>` `C=5/4`
`x=-1` then `5=4A` `=>` `A=5/4`
`x=0` `=>` `5=A+B+C-D`
`B-D=5-5/4-5/4=5/2`
Coefficients of `x^3` then `0=B+D`
`B=5/4` and `D=-5/4`
So `=A/(x+1)^2+B/(x+1)+C/(x-1)^2+D/(x-1)`
`=5/(4(x+1)^2)+5/(4(x+1))+5/(4(x-1)^2)-5/(4(x-1))`