Question

Resolve the following rational expressions into partial fractions

`1/(x^4 - 1)`

Answer 1

`1/(x^4 - 1) = 1/((x^2)^2 - 1^2`

= `1/((x^2 + 1)(x^2 - 1))`

= `1/((x^2 + 1)(x + 1)(x - 1))`

`1/(x^4 - 1) = ("A"x + "B")/(x^2 + 1) + "C"/(x + 1) + "D"/(x - 1)`

`1/(x^4 - 1) = (("A"x + "B")(x + 1)(x - 1) + "C"(x^2 + 1)(x - 1) + "D"(x + 1)(x^2 + 1))/((x^2 + 1)(x + 1)(x - 1))`

1 = Ax(x + 1)(x – 1) + B(x + 1)(x – 1) + C(x<sup2 + 1)(x – 1) + D(x + 1)(x² + 1)   ......(1)

Put x = 1 in equation (1)

1 = A(1)(1 + 1)(1 – 1) + B(1 + 1)(1 – 1) + C(1<sup2 + 1)(1 – 1) + D(1 + 1)(1<sup2 + 1)

1 = A × 0 + B × 0 + C × 0 + D(2)(2)

⇒ 1 = – 4D

⇒ D = `1/4`

Put x = – 1 in equation (1)

1 = A(– 1)(– 1 + 1)(– 1 – 1) + B(– 1 + 1)(– 1 – 1) + C((– 1)² + 1)(– 1 + 1) + D(– 1 + 1)((– 1)² + 1)

1 = A × 0 + B × 0 + C(2)(– 2) + D × 0

⇒ 1 = – 4C

⇒ C = `- 1/4`

Put x = 0 in equation (1)

I = A(0)(0 + 1)(0 – 1) + B(0 + 1)(0 – 1) + C(0² + 1)(0 – 1) + D(0 + 1)(0² + 1)

1 = A × O + B(– 1) + C(– 1) + D(1)

⇒ 1 = – B – C + D

1 = `- "B" + 1/4 + 1/4`

⇒ B = `1/2 - 1 = - 1/2`

⇒ B = `- 1/4`
In equation (1), equate the coefficient of x³ on both sides

0 = A + C + D

⇒ 0 = `"A" - 1/4 + 1/4`

⇒ A = 0

∴ The required partial fraction is

`1/(x^4 - 1) = (0^x - 1/2)/(x^2 + 1) + (- 1/4)/(x + 1) + (1/4)/(x - 1)`

`1/(x^4 - 1) = - 1/(2(x^2 + 1)) - 1/(4(x + 1)) + 1/(4(x - 1))`