Solve into partial fraction for (x + 2)/(x - 1)(x + 3)^2

Question

Solve into partial fraction for \(\frac{x+2}{(x-1)(x+3)^2}\)

Answer 1

Here the denominator has three factors. So given fraction can be expressed as a sum of three simple fractions.

Let \(\frac{x+2}{(x-1)(x+3)^2}\) = \(\frac{A}{x-1}+\frac{B}{x+3}+\frac{C}{(x+3)^2}\) ... (1)

Multiply both sides by (x – 1) (x + 3)² we get 

\(\frac{x+2}{(x-1)(x+3)^2}\) (x – 1) (x + 3)² = \(\frac{A}{x-1}\) (x – 1) (x + 3)² + \(\frac{B}{x+3}\) (x – 1) (x + 3)² + \(\frac{C}{(x+3)^2}\) (x – 1) (x + 3)²

x + 2 = A(x + 3)² + B(x – 1) (x + 3) + C(x – 1) … (2) 

Put x = 1 in (2) we get

1 + 2 = A(1 + 3)² + 0 + 0 

3 = A(4)² 

A = \(\frac{3}{16}\)

Put x = -3 in (2) we get 

-3 + 2 = 0 + 0 + C(-3 – 1) 

-1 = C(-4) 

C = \(\frac{1}{4}\)

Comparing coefficient of x² on both sides of (2) we get, 

0 = A + B 

0 = \(\frac{3}{16}\) + B 

B = \(-\frac{3}{16}\)

Using A = \(\frac{3}{16}\), B = \(-\frac{3}{16}\), C = \(\frac{1}{4}\) in (1) we get,

\(\frac{x+2}{(x-1)(x+3)^2}\)