# Math Help - Partial Fractions

1. ## Partial Fractions

Hi

How do i solve this partial fraction:

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

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

$\frac{4x^2 - 2x}{(x + 1)(x^2 +1)} = \frac{A}{x + 1} + \frac{B}{x^2 + 1}$
$\frac{A(x^2 + 1) + B(x+1)}{(x+1)(x^2+1)}$
$4x^2 = 2x = A(x^2 +1) + B(x+1)$
Let x = -1
4+2 = 2A
6 = 2A
A = 3
$4x^2 - 2x = Ax^2 + A + Bx + B$
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?

2. Shouldn't the second numerator be of the form "Bx + C", due to the fact that the denominator is of degree 2...?

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