Solve into partial fraction for 1/(x - 1)(x + 2)^2

Question

Solve into partial fraction for \(\frac{1}{(x-1)(x+2)^2}\)

Answer 1

Here the denominator has repeated factors. So we write

\(\frac{1}{(x-1)(x+2)^2}\) = \(\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}...(1)\)

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 = \(\frac{1}{9}\)

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 = \(\frac{-1}{3}\)

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 = \(\frac{1}{9}\) (∴ A = \(\frac{1}{9}\))

B = \(-\frac{1}{9}\)

Using A = \(\frac{1}{9}\), B = \(-\frac{1}{9}\), C = \(\frac{-1}{3}\) in (1) we get,