frac{x^2 + 1}{(2x - 1)(x^2 - 1)} = | vedclass

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Question

(B) We express the given fraction as partial fractions: $\frac{x^2 + 1}{(2x - 1)(x^2 - 1)} = \frac{A}{2x - 1} + \frac{B}{x + 1} + \frac{C}{x - 1}$ Mul

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We express the given fraction as partial fractions: $\frac{x^2 + 1}{(2x - 1)(x^2 - 1)} = \frac{A}{2x - 1} + \frac{B}{x + 1} + \frac{C}{x - 1}$ Multiplying both sides by $(2x - 1)(x^2 - 1)$,we get: $x^2 + 1 = A(x^2 - 1) + B(2x - 1)(x - 1) + C(2x - 1)(x + 1)$ For $x = 1$: $1^2 + 1 = C(2(1) - 1)(1 + 1)$ $\Rightarrow 2 = 2C$ $\Rightarrow C = 1$ For $x = -1$: $(-1)^2 + 1 = B(2(-1) - 1)(-1 - 1)$ $\Rightarrow 2 = B(-3)(-2)$ $\Rightarrow 2 = 6B$ $\Rightarrow B = \frac{1}{3}$ For $x = \frac{1}{2}$: $(\frac{1}{2})^2 + 1 = A((\frac{1}{2})^2 - 1)$ $\Rightarrow \frac{5}{4} = A(\frac{1}{4} - 1)$ $\Rightarrow \frac{5}{4} = -\frac{3}{4}A$ $\Rightarrow A = -\frac{5}{3}$ Substituting the values of $A, B,$ and $C$: $\frac{-5}{3(2x - 1)} + \frac{1}{3(x + 1)} + \frac{1}{x - 1}$

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

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