Finding the Partial Fractions Using Coefficients

Question

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

For the above question when I find the partial fractions using the elimination method I got the following answer.

(3-2x)/(x+1)(1-x)^2 = 5/4(x+1) - 5/4(1-x) + 1/2(1-x)^2

But When I tried to find the partial fractions using the coefficient method I got a different answer. That answer is this:

(3-2x)/(x+1)(1-x)^2 = 5/4(x+1) + 5/4(1-x) + 1/2(1-x)^2

However, the correct answer is the answer I observed from the elimination method.

I have stated the calculations I have done when using the coefficient method. Can someone please point me out what is wrong with my calculations and why I got a different answer?

(3-2x)/(x+1)(1-x)^2 = A(x+1) + B/(1-x) + C/(1-x)^2

(3-2x) = A(1-x)^2 + B(x+1)(1-x) + C(x+1)

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

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

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

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

        
let x =1;

x^2: 0 = (A-B) --------------- 1

x: -2 = -2A+c  --------------- 2

C: 3 = A +B +C --------------- 3


1 + 3:
      3 = 2A +C --------------- 4


2 + 4:
      C = 1/2

C= (1/2) substitute in 2:

                        -2 = -2A + (1/2)

                         A = 5/4

A = (5/4) substitute in 1:

                        0 = (5/4) - B

                        B = 5/4


Thus ;

      (3-2x)/(x+1)(1-x)^2 = 5/4(x+1) + 5/4(1-x) + 1/2(1-x)^2

Answer 1

There is nothing wrong in your computations. The partial fraction decomposition is$$\frac{3-2x}{(x+1)(x-1)^2}=\frac{5/4}{x+1}+\frac{5/4}{1-x}+\frac{1/2}{(1-x)^2}$$and the other answer is wrong.