# Thread: Quick Question on Partial Fractions Problem

1. ## Quick Question on Partial Fractions Problem

Hi
Can someone tell me if my answer is correct or my book's answers

1) Resolve into a partial fraction: $\frac{x+1}{(x^2+4)(x^2-1)}$

i got $\frac{\frac{1}{5}}{(x-1)}-\frac{\frac{1}{5}}{(x^2+4)}$

however book's answers is $\frac{\frac{1}{5}}{(x-1)}-\frac{(x+1)}{5(x^2+4)}
$

P.S

2. Originally Posted by Paymemoney
Hi
Can someone tell me if my answer is correct or my book's answers

1) Resolve into a partial fraction: $\frac{x+1}{(x^2+4)(x^2-1)}$

i got $\frac{\frac{1}{5}}{(x-1)}-\frac{\frac{1}{5}}{(x^2+4)}$

however book's answers is $\frac{\frac{1}{5}}{(x-1)}-\frac{(x+1)}{5(x^2+4)}
$

P.S

$\frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{Ax + B}{(x^2 + 4)} + \frac{Cx + D}{(x^2 - 1)}$

$= \frac{(Ax + B)(x^2 - 1) + (Cx + D)(x^2 + 4)}{(x^2 + 4)(x^2 - 1)}$.

So $(Ax + B)(x^2 - 1) + (Cx + D)(x^2 + 4) = x + 1$

$Ax^3 - Ax + Bx^2 - B + Cx^3 + 4Cx + Dx^2 + 4D = x + 1$

$(A + C)x^3 + (B + D)x^2 + (4C - A)x + 4D - B = 0x^3 + 0x^2 + 1x + 1$.

So $A + C = 0$

$B + D = 0$

$4C - A = 1$

$4D - B = 1$

Solving these equations simultaneously gives

$A = -\frac{1}{5}, B = -\frac{1}{5}, C = \frac{1}{5}, D = \frac{1}{5}$.

So $\frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{-\frac{1}{5}x - \frac{1}{5}}{x^2 + 4} + \frac{\frac{1}{5}x + \frac{1}{5}}{x^2 - 1}$

$= -\frac{x + 1}{5(x^2 + 4)} + \frac{x + 1}{5(x^2 - 1)}$

$= \frac{x + 1}{5(x - 1)(x + 1)} - \frac{x + 1}{5(x^2 + 4)}$

$= \frac{1}{5(x - 1)} - \frac{x + 1}{5(x^2 + 4)}$.

3. Originally Posted by Prove It
$\frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{Ax + B}{(x^2 + 4)} + \frac{Cx + D}{(x^2 - 1)}$
yeh i made it $= \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 + 4)}$.
yeh i made it $= \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 + 4)}$.
$\frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 4}$.