# How do I find the partial fraction decomposition of (x^4)/(x^4-1) ?

Sep 18, 2014

By Partial Fraction Decomposition, we can write

${x}^{4} / \left\{{x}^{4} - 1\right\} = 1 - \frac{\frac{1}{4}}{x + 1} + \frac{\frac{1}{4}}{x - 1} - \frac{\frac{1}{2}}{{x}^{2} + 1}$.

Let us look at some details.

By rewriting a bit,

$\frac{{x}^{4}}{{x}^{4} - 1} = 1 + \frac{1}{{x}^{4} - 1}$

Let us find the partial fractions of

$\frac{1}{{x}^{4} - 1}$

by factoring out the denominator,

$= \frac{1}{\left(x + 1\right) \left(x - 1\right) \left({x}^{2} + 1\right)}$

by splitting into the partial fraction form,

$= \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C x + D}{{x}^{2} + 1}$

by taking the common denominator,

$= \frac{A \left(x - 1\right) \left({x}^{2} + 1\right) + B \left(x + 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) \left({x}^{2} - 1\right)}{\left(x + 1\right) \left(x - 1\right) \left({x}^{2} + 1\right)}$

by simplifying the numerator,

$= \frac{\left(A + B + C\right) {x}^{3} + \left(- A + B + D\right) {x}^{2} + \left(A + B - C\right) x + \left(- A + B - D\right)}{{x}^{4} - 1}$

Since the numerator is originally 1, by matching the coefficients,

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

(5) $2 A + 2 B = 0$

(6) $- 2 A + 2 B = 1$

(7) $B = \frac{1}{4}$

By plugging (7) into (5),

(8) $A = - \frac{1}{4}$

By plugging (7) and (8) into (1),

(9) $C = 0$

By plugging (7) and (8) into (2),

(10) $D = - \frac{1}{2}$

By (5), (6), (9), and (10),

${x}^{4} / \left\{{x}^{4} - 1\right\} = 1 - \frac{\frac{1}{4}}{x + 1} + \frac{\frac{1}{4}}{x - 1} - \frac{\frac{1}{2}}{{x}^{2} + 1}$