# Partial fraction decomposition.

• May 18th 2010, 11:00 PM
ChristopherDunn
Partial fraction decomposition.
Decompose into partial fractions: $\displaystyle \frac{x}{x^4-a^4}$
• May 18th 2010, 11:09 PM
Prove It
Quote:

Originally Posted by ChristopherDunn
Decompose into partial fractions: $\displaystyle \frac{x}{x^4-a^4}$

$\displaystyle \frac{x}{x^4 - a^4} = \frac{x}{(x^2)^2 - (a^2)^2}$

$\displaystyle = \frac{x}{(x^2 - a^2)(x^2 + a^2)}$

$\displaystyle = \frac{x}{(x - a)(x + a)(x^2 + a^2)}$.

Now try using partial fractions with

$\displaystyle \frac{A}{x - a} + \frac{B}{x + a} + \frac{Cx + D}{x^2 + a^2} = \frac{x}{(x - a)(x + a)(x^2 + a^2)}$.
• May 19th 2010, 06:45 PM
ChristopherDunn
1. $\displaystyle \frac{A}{x - a} + \frac{B}{x + a} + \frac{Cx + D}{x^2 + a^2} = \frac{x}{(x - a)(x + a)(x^2 + a^2)}$
2. $\displaystyle x=A(x+a)(x^2+a^2)+B(x-a)(x^2+a^2)+(Cx+D)(x-a)(x+a)$

let x=a;
1. $\displaystyle a=A(a+a)(a^2+a^2)$
2. $\displaystyle a=4a^3A$
3. $\displaystyle A=\frac{1}{4a^2}$

let x=-a;
1. $\displaystyle -a=B(-a-a)((-a)^2+a^2)$
2. $\displaystyle -a=-4a^3B$
3. $\displaystyle B=\frac{1}{4a^2}$

How do I find C and D?
• May 19th 2010, 08:02 PM
Prove It
Quote:

Originally Posted by ChristopherDunn
1. $\displaystyle \frac{A}{x - a} + \frac{B}{x + a} + \frac{Cx + D}{x^2 + a^2} = \frac{x}{(x - a)(x + a)(x^2 + a^2)}$
2. $\displaystyle x=A(x+a)(x^2+a^2)+B(x-a)(x^2+a^2)+(Cx+D)(x-a)(x+a)$

let x=a;
1. $\displaystyle a=A(a+a)(a^2+a^2)$
2. $\displaystyle a=4a^3A$
3. $\displaystyle A=\frac{1}{4a^2}$

let x=-a;
1. $\displaystyle -a=B(-a-a)((-a)^2+a^2)$
2. $\displaystyle -a=-4a^3B$
3. $\displaystyle B=\frac{1}{4a^2}$

How do I find C and D?

Let $\displaystyle x = 0$ to find $\displaystyle D$.

Then you can use your information about $\displaystyle A, B, D$ to find $\displaystyle C$.
• May 19th 2010, 08:50 PM
ChristopherDunn
That'll do it -- thanks!