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

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Answer 1 · 2014-09-18T21:42:37

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}#