#1**0 **

Do you mean partial fractions?

\(\dfrac{1}{x^2 - 1}=\dfrac{a}{x-1}+\dfrac{b}{x+1}\\ 1=a(x+1)+b(x-1)=(a+b)x+a-b \\a=1/2\text{ and }b=-1/2\\ \therefore \dfrac{1}{x^2-1}=\dfrac{1}{2(x-1)}-\dfrac{1}{2(x+1)}\)

We made 1/(x^2-1) seperate into 2 fractions: 1/(2(x-1)) and -1/(2(x+1)).

MaxWong
Aug 15, 2017