Question

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

Answer 1

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}`