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

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(D) Let $\frac{x^2 + 1}{(2x - 1)(x - 1)(x + 1)} = \frac{A}{2x - 1} + \frac{B}{x - 1} + \frac{C}{x + 1}$.Multiplying both sides by $(2x - 1)(x - 1)(x +

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(D) Let $\frac{x^2 + 1}{(2x - 1)(x - 1)(x + 1)} = \frac{A}{2x - 1} + \frac{B}{x - 1} + \frac{C}{x + 1}$.Multiplying both sides by $(2x - 1)(x - 1)(x + 1)$,we get:$x^2 + 1 = A(x - 1)(x + 1) + B(2x - 1)(x + 1) + C(2x - 1)(x - 1)$.For $x = 1/2$: $(1/4) + 1 = A(1/2 - 1)(1/2 + 1) \implies 5/4 = A(-1/2)(3/2) \implies 5/4 = -3/4 A \implies A = -5/3$.For $x = 1$: $1 + 1 = B(2(1) - 1)(1 + 1) \implies 2 = B(1)(2) \implies 2 = 2B \implies B = 1$.For $x = -1$: $(-1)^2 + 1 = C(2(-1) - 1)(-1 - 1) \implies 2 = C(-3)(-2) \implies 2 = 6C \implies C = 1/3$.Thus,the partial fraction is $\frac{-5}{3(2x - 1)} + \frac{1}{(x - 1)} + \frac{1}{3(x + 1)}$.

Resolve $\frac{x^2 + 1}{(2x - 1)(x^2 - 1)}$ into partial fractions.

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