# Partial fraction decomposition.

#### ChristopherDunn

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

#### Prove It

MHF Helper
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

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

Rapha

#### 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?

Rapha

#### Prove It

MHF Helper
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 $$\displaystyle C$$.

ChristopherDunn

#### ChristopherDunn

That'll do it -- thanks!