Question

Resolve into partial fraction for the following:

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

Answer 1

Here the denominator has repeated factors. So we write

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

Multiply both sides by (x – 1) (x + 2)² we get

1 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1) …… (2)

Put x = 1 in (2) we get

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

1 = A(3²) + 0 + 0

1 = 9A

A = `1/9`

Put x = -2 in (2) we get

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

1 = 0 + 0 + C(-3)

C = `(-1)/3`

From (2) we have

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

0x² + 1 = A(x² + 4x + 4) + B(x² + x – 2) + C(x – 1)

Equating coefficient of x² on both sides we get

0 = A + B

0 = `1/9 (therefore "A" = 1/9)`

B = `- 1/9`

Using A = `1/9`, B = `- 1/9`, C = `- 1/3` in (1) we get

`1/((x - 1)(x + 2)(x + 2)^2) = (1/9)/(x - 1) + (-1/9)/(x + 2) + (-1/3)/(x + 2)^2`

`= 1/(9(x - 1)) - 1/(9(x + 2)) - 1/(3(x + 2)^2)`