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

Question

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

Sadie Ackerman · Accepted Answer

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

Sadie Alexander · Answer

undefined To write the partial fraction decomposition of the rational expression $ \frac{5}{(x^2-1)^2} $:

1. Factor the denominator $ (x^2-1)^2 $.
2. Express $ \frac{5}{(x^2-1)^2} $ as a sum of partial fractions, with each partial fraction having a denominator corresponding to one of the factors of $ (x^2-1)^2 $.

The factorization of $ (x^2-1)^2 $ is $ (x-1)^2(x+1)^2 $.

Therefore, the partial fraction decomposition is:

$$ \frac{5}{(x^2-1)^2} = \frac{A}{(x-1)^2} + \frac{B}{(x-1)} + \frac{C}{(x+1)^2} + \frac{D}{(x+1)} $$

Where $ A $, $ B $, $ C $, and $ D $ are constants to be determined.