# Quick Question on Partial Fractions Problem

• Mar 26th 2010, 05:45 AM
Paymemoney
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
• Mar 26th 2010, 06:00 AM
Prove It
Quote:

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

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

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

$= \frac{1}{5(x - 1)} - \frac{x + 1}{5(x^2 + 4)}$.
• Mar 26th 2010, 03:30 PM
Paymemoney
Quote:

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 $= \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 + 4)}$.
• Mar 26th 2010, 10:49 PM
Prove It
Quote:

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

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