Hi

How do i solve this partial fraction:

$\displaystyle \frac{4x^2 - 2x}{(x + 1)(x^2 +1)}$

I tried using the same method as I did before...

$\displaystyle \frac{4x^2 - 2x}{(x + 1)(x^2 +1)} = \frac{A}{x + 1} + \frac{B}{x^2 + 1}$

$\displaystyle \frac{A(x^2 + 1) + B(x+1)}{(x+1)(x^2+1)}$

$\displaystyle 4x^2 = 2x = A(x^2 +1) + B(x+1)$

Let x = -1

4+2 = 2A

6 = 2A

A = 3

$\displaystyle 4x^2 - 2x = Ax^2 + A + Bx + B$

Differentiate both sides.

8x - 2 = 2Ax + B

Substitute in A = 3

8x - 2 = 6x + B

2x-2 = B

I keep on getting 2x - 2 as the value for B even though it's meant to be x - 3.

Could someone please point out what i did wrong?