If frac{x}{(x - 1)(x^2 + 1)^2} = frac{1}{4} l | vedclass

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(A) Given the equation: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4} \left[ \frac{1}{x - 1} - \frac{x + 1}{x^2 + 1} \right] + y$We express the partial

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

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(A) Given the equation: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4} \left[ \frac{1}{x - 1} - \frac{x + 1}{x^2 + 1} \right] + y$We express the partial fraction decomposition of the left side as: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} + \frac{Dx + E}{(x^2 + 1)^2}$Comparing the given expression with the standard decomposition,we identify $y$ as the term $\frac{Dx + E}{(x^2 + 1)^2}$.Multiplying by the common denominator $(x - 1)(x^2 + 1)^2$:$x = \frac{1}{4}(x^2 + 1)^2 - \frac{1}{4}(x + 1)(x - 1)(x^2 + 1) + y(x - 1)(x^2 + 1)^2$Solving for the coefficients,we find $y = \frac{1 - x}{2(x^2 + 1)^2}$.

If $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4} \left[ \frac{1}{x - 1} - \frac{x + 1}{x^2 + 1} \right] + y$,then $y =$

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