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

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(A) We use the method of partial fractions for the expression $\frac{x}{(x-1)(x^2+1)^2}$.Let $\frac{x}{(x-1)(x^2+1)^2} = \frac{A}{x-1} + \frac{Bx+C}{x

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$\frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$

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(A) We use the method of partial fractions for the expression $\frac{x}{(x-1)(x^2+1)^2}$.Let $\frac{x}{(x-1)(x^2+1)^2} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}$.Multiplying by $(x-1)(x^2+1)^2$,we get $x = A(x^2+1)^2 + (Bx+C)(x-1)(x^2+1) + (Dx+E)(x-1)$.Setting $x=1$,we get $1 = A(2)^2$,so $A = \frac{1}{4}$.Comparing coefficients or substituting values for $x$,we find $B = -\frac{1}{4}$,$C = -\frac{1}{4}$,$D = -\frac{1}{2}$,and $E = \frac{1}{2}$.Substituting these values,we have $\frac{x}{(x-1)(x^2+1)^2} = \frac{1}{4(x-1)} - \frac{x+1}{4(x^2+1)} + \frac{-x/2 + 1/2}{(x^2+1)^2}$.This simplifies to $\frac{1}{4}\left[\frac{1}{x-1} - \frac{x+1}{x^2+1}\right] + \frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$.Comparing this with the given equation,we get $y = \frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$.

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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