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

Question

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

Answer 1

Let \(\frac{x-2}{(x+2)(x-1)^2}\)

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

Put x = 1 in (2) we get 

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

-1 = 0 + 0 + 3C 

C = \(-\frac{1}{3}\)

Put x = -2 in (2) we get

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

-4 = A(-3)² + 0 + 0 

-4 = 9A 

A = \(\frac{-4}{9}\)

From (2) we have, 

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

Equating coefficients of x² on both sides we get 

0 = A + B 

0 = \(\frac{-4}{9}\) + B (∵ A = \(\frac{-4}{9}\)) 

B = \(\frac{4}{9}\)

Using A = \(\frac{-4}{9}\), B = \(\frac{4}{9}\), C = \(-\frac{1}{3}\) in (1) we get