Question-67561

Question Number 67561 by azizullah last updated on 28/Aug/19

Answered by Rasheed.Sindhi last updated on 28/Aug/19

L e t \frac{2 x + 1}{(x^{2} + 1) (x - 1)} = \frac{A x + B}{x^{2} + 1} + \frac{C}{x - 1}

(A x + B) (x - 1) + C (x^{2} + 1) = 2 x + 1

x = 1 \Rightarrow 2 C = 3 \Rightarrow C = 3 / 2

{A x}^{2} + B x - A x - B + {C x}^{2} + C = 2 x + 1

(A + C) x^{2} + (B - A) x + C - B = 0 x^{2} + 2 x + 1

A + C = 0 \Rightarrow A = - C = - 3 / 2

B - A = 2 \Rightarrow B = A + 2 = - 3 / 2 + 2 = 1 / 2

C - B = 1 \Rightarrow 3 / 2 - 1 / 2 = 1

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

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

Commented by azizullah last updated on 29/Aug/19

Sir Alot of thanks!

Commented by Rasheed.Sindhi last updated on 30/Aug/19

You're welcome mr Azizullah!