2x+1 ÷ (x-1) (x-2) (x-3) resolve into partial fraction

Question

\(\frac{2x+1}{(x-1)(x-2)(x-3)}\) resolve into partial fraction.

Answer 1

\(\frac{2x+1}{(x-1)(x-2)(x-3)}=\frac A{x-1}+\frac B{x-2}+\frac c{x-3}\) 

⇒ 2x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + c (x - 1)(x - 2)---(1)

Put  x = 1 in equation (1) we get

2A = 2 + 1 = 3 ⇒ A = 3/2

Put x = 2 in equation (1), we get

-B = 5 ⇒ B = -5

Put x = 3 in equation (1), we get

2c = 7 ⇒ c = 7/2

Hence, \(\frac{2x+1}{(x-1)(x-2)(x-3)}=\frac{3/2}{x-1}-\frac5{x-2}+\frac{7/2}{x-3}\)