Answer:
[tex]\frac{1}{x-1}[/tex] - [tex]\frac{1}{x+2}[/tex] - [tex]\frac{3}{(x+2)^2}[/tex]
Step-by-step explanation:
Expressing as a partial fraction
[tex]\frac{9}{(x-1)(x+2)^2}[/tex] = [tex]\frac{A}{(x-1)}[/tex] + [tex]\frac{B}{(x+2)}[/tex] + [tex]\frac{C}{(x+2)^2}[/tex] ( A, B, C are numerical values to be found )
Multiply both sides by (x - 1)(x + 2)²
9 = A(x + 2)² + B(x - 1)(x + 2) + C(x - 1)
Choose value for x that are zeros of the factors and substitute into right side
x = - 2 : 9 = - 3C ⇒ C = - 3
x = 1 : 9 = 9A ⇒ A = 1
x = 0 : 9 = 4A - 2B - C , that is
9 = 4 - 2B + 3 = - 2B + 7 ( subtract 7 from both sides )
2 = - 2B ⇒ B = - 1, then
A = 1, B = - 1, C = - 3
Thus
[tex]\frac{9}{(x-1)(x+2)^2}[/tex] = [tex]\frac{1}{x-1}[/tex] - [tex]\frac{1}{x+2}[/tex] - [tex]\frac{3}{(x+2)^2}[/tex]
