Resolve into partial fractions x/(x^2-1)(x-1)?

Answer 1

x / [(x^2-1)(x-1)]

x / [(x+1)(x-1)(x-1)] expanded (x^2 - 1)

x / [(x+1)(x-1)^2)] combined like terms

x / [(x+1)(x-1)^2)] = [A / (x+1)] + [B / (x-1)] + [C / (x-1)^2]

The next step is to multiply both sides of the above equation by [(x+1)(x-1)^2].

x = A(x-1)^2 + B(x+1)(x-1) + C(x+1)

x = Ax^2 + Bx^2 - 2Ax + Cx + A - B + C expanded

x = (A+B)x^2 + (-2A+C)x + A - B + C

A+B=0
-2A+C=1
A-B+C=0

Solve the system of equations to get:

A = -1/4
B = 1/4
C = 1/2

Resubstitute A, B, and C into the partial fractions:

-1 / [4(x+1)] + 1 / [4(x-1)] + 1 / [2(x-1)^2]

There are four different cases for expressing a rational function into partial fractions. Your math textbook should have them or check the internet for them. Good luck!

Answer 2

x/(x^2-1)(x-1) =x/ x^3 -x^2 -x +1

Answer 3

the given answer is wrong u can check out in any of the text book there u see under the category of similar partial fractions.. sory i forgot