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Question: \[\frac{x^{2} + 1}{(2x - 1)(x^{2} - 1)} =\]...

x2+1(2x−1)(x2−1)=\frac{x^{2} + 1}{(2x - 1)(x^{2} - 1)} =

A

(a) \frac{- 5}{3(2x - 1)} + \frac{3}{(x + 1)} + \frac{1}{(x - 1)}

B

\frac{- 5}{3(2x - 1)} + \frac{1}{3(x + 1)} + \frac{1}{(x - 1)}

C

\frac{1}{2x - 1} + \frac{5}{(x + 1)} - \frac{3}{(x - 1)}

D

None of these

Answer

\frac{- 5}{3(2x - 1)} + \frac{1}{3(x + 1)} + \frac{1}{(x - 1)}

Explanation

Solution

x2+1(2x−1)(x2−1)=A(2x−1)+Bx+1+Cx−1\frac{x^{2} + 1}{(2x - 1)(x^{2} - 1)} = \frac{A}{(2x - 1)} + \frac{B}{x + 1} + \frac{C}{x - 1}

⇒ x2+1=A(x2−1)+B(2x−1)(x−1)+C(x+1)(2x−1)x^{2} + 1 = A(x^{2} - 1) + B(2x - 1)(x - 1) + C(x + 1)(2x - 1)

For x=1,x = 1, 2=2C⇒C=12 = 2C \Rightarrow C = 1

For x=−1x = - 1, 2=6B⇒B=132 = 6B \Rightarrow B = \frac{1}{3}

For x=12x = \frac{1}{2}, 54=−34A⇒A=−53\frac{5}{4} = - \frac{3}{4}A \Rightarrow A = - \frac{5}{3}

∴\therefore Given expression = −531(2x−1)+131x+1+1x−1- \frac{5}{3}\frac{1}{(2x - 1)} + \frac{1}{3}\frac{1}{x + 1} + \frac{1}{x - 1}