# 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
The book's answer is correct.

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

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

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
The book's answer is correct.

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

4. Originally Posted by Paymemoney
yeh i made it $= \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 + 4)}$.
When dealing with partial fractions, it's always easiest to work with two factors, and then break it down further later if possible.

Also, your way wouldn't work as it should actually be

$\frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 4}$.