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

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(A) Given the equation: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4}\left[ \frac{1}{(x - 1)} - \frac{x + 1}{x^2 + 1} \right] + y$First,simplify the ter

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

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(A) Given the equation: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4}\left[ \frac{1}{(x - 1)} - \frac{x + 1}{x^2 + 1} \right] + y$First,simplify the term inside the bracket: $\frac{1}{x - 1} - \frac{x + 1}{x^2 + 1} = \frac{(x^2 + 1) - (x^2 - 1)}{(x - 1)(x^2 + 1)} = \frac{2}{(x - 1)(x^2 + 1)}$Substituting this back: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4} \cdot \frac{2}{(x - 1)(x^2 + 1)} + y = \frac{1}{2(x - 1)(x^2 + 1)} + y$Solving for $y$: $y = \frac{x}{(x - 1)(x^2 + 1)^2} - \frac{1}{2(x - 1)(x^2 + 1)}$Taking common denominator $2(x - 1)(x^2 + 1)^2$: $y = \frac{2x - (x^2 + 1)}{2(x - 1)(x^2 + 1)^2} = \frac{-(x^2 - 2x + 1)}{2(x - 1)(x^2 + 1)^2} = \frac{-(x - 1)^2}{2(x - 1)(x^2 + 1)^2} = \frac{-(x - 1)}{2(x^2 + 1)^2} = \frac{1 - x}{2(x^2 + 1)^2}$

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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