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

1 Answer
Sep 18, 2014

By Partial Fraction Decomposition, we can write

x^4/{x^4-1}=1-{1/4}/{x+1}+{1/4}/{x-1}-{1/2}/{x^2+1}.

Let us look at some details.

By rewriting a bit,

{x^4}/{x^4-1}=1+1/{x^4-1}

Let us find the partial fractions of

1/(x^4-1)

by factoring out the denominator,

=1/{(x+1)(x-1)(x^2+1)}

by splitting into the partial fraction form,

=A/{x+1}+B/{x-1}+{Cx+D}/{x^2+1}

by taking the common denominator,

={A(x-1)(x^2+1)+B(x+1)(x^2+1)+(Cx+D)(x^2-1)}/{(x+1)(x-1)(x^2+1)}

by simplifying the numerator,

={(A+B+C)x^3+(-A+B+D)x^2+(A+B-C)x+(-A+B-D)}/{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

By adding (1) and (3),

(5) 2A+2B=0

By adding (2) and (4),

(6) -2A+2B=1

By adding (5) and (6),

(7) B=1/4

By plugging (7) into (5),

(8) A=-1/4

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

(9) C=0

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

(10) D=-1/2

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

x^4/{x^4-1}=1-{1/4}/{x+1}+{1/4}/{x-1}-{1/2}/{x^2+1}