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

1.	Solve into partial fraction for \(\frac{4x+1}{(x-2)(x+1)}\)
Answer» 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 \(\frac{4x+1}{(x-2)(x+1)}\) = \(\frac{3}{x-2}+\frac{1}{x+1} \)

Answer»

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

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

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