Solve into partial fraction for (4x + 1)/(x - 2)(x + 1)

Question

Solve into partial fraction for \(\frac{4x+1}{(x-2)(x+1)}\)

Answer 1

Let \(\frac{4x+1}{(x-2)(x+1)}\) = \(\frac{A}{x-2}+\frac{B}{x+1} ...(1)\)

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

4x + 1 = A(x + 1) + B(x – 2) … (2)

Put x = -1 in (2) we get

4(-1) + 1 = A(-1 + 1) + B(-1 – 2)

-4 + 1 = A(0) + B(-3)

-3 = B(-3)

B = \(\frac{-3}{-3}\) = 1

Put x = 2 in (2) we get

4(2) + 1 = A(2 + 1) + B(2 – 2)

8 + 1 = A(3) + B(0)

9 = 3A

A = 3

Using A = 3, B = 1 in (1) we get

\(\frac{4x+1}{(x-2)(x+1)}\) = \(\frac{3}{x-2}+\frac{1}{x+1} \)