How do you write the partial fraction decomposition of the rational expression 5/(x^2-1)^25(x2−1)2?

1 Answer
Narad T.
Nov 1, 2016

The answer is =5/(4(x+1)^2)+5/(4(x+1))+5/(4(x-1)^2)-5/(4(x-1))=54(x+1)2+54(x+1)+54(x−1)2−54(x−1)

Explanation:

Let's factorise the denominator (x^2-1)^2=(x+1)(x-1(x+1)(x-1)(x2−1)2=(x+1)(x−1(x+1)(x−1)
So 5/(x^2-1)^2=5/((x+1)^2(x-1)^2)5(x2−1)2=5(x+1)2(x−1)2
=A/(x+1)^2+B/(x+1)+C/(x-1)^2+D/(x-1)=A(x+1)2+Bx+1+C(x−1)2+Dx−1
5=5=A(x-1)^2+B(x+1)(x-1)^2+C(x+1)^2+D(x-1)(x+1)^2A(x−1)2+B(x+1)(x−1)2+C(x+1)2+D(x−1)(x+1)2
Let x=1x=1 then 5=4C5=4C=>⇒C=5/4C=54
x=-1x=−1 then 5=4A5=4A=>⇒A=5/4A=54
x=0 x=0=>⇒ 5=A+B+C-D5=A+B+C−D
B-D=5-5/4-5/4=5/2B−D=5−54−54=52
Coefficients of x^3x3 then 0=B+D0=B+D
B=5/4B=54 and D=-5/4D=−54
So =A/(x+1)^2+B/(x+1)+C/(x-1)^2+D/(x-1)=A(x+1)2+Bx+1+C(x−1)2+Dx−1
=5/(4(x+1)^2)+5/(4(x+1))+5/(4(x-1)^2)-5/(4(x-1))=54(x+1)2+54(x+1)+54(x−1)2−54(x−1)

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