The partial fractions of frac{x^2}{(x - 1)^3( | vedclass

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Question

(C) Let $x - 1 = y$,so $x = y + 1$.Then,$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)} = -\frac{1 + 2y

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$\frac{-1}{(x - 1)^3} + \frac{-3}{(x - 1)^2} + \frac{-4}{(x - 1)} + \frac{4}{(x - 2)}$

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(C) Let $x - 1 = y$,so $x = y + 1$.Then,$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)} = -\frac{1 + 2y + y^2}{y^3(1 - y)}$.Dividing $(1 + 2y + y^2)$ by $(y - 1)$ gives $(y^2 + 2y + 1) = (y - 1)(y + 3) + 4$.Thus,$\frac{y^2 + 2y + 1}{y^3(y - 1)} = \frac{(y - 1)(y + 3) + 4}{y^3(y - 1)} = \frac{y + 3}{y^3} + \frac{4}{y^3(y - 1)}$.Using partial fractions for $\frac{4}{y^3(y - 1)} = \frac{A}{y} + \frac{B}{y^2} + \frac{C}{y^3} + \frac{D}{y - 1}$,we find $D = 4$,$C = -4$,$B = -4$,$A = -4$.Substituting back,the expression becomes $\frac{-1}{(x - 1)^3} + \frac{-3}{(x - 1)^2} + \frac{-4}{(x - 1)} + \frac{4}{(x - 2)}$.

The partial fractions of $\frac{x^2}{(x - 1)^3(x - 2)}$ are

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