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

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(C) Let $(x - 1) = y$,then $x = y + 1$.Substituting this into the expression:$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 +

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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 $(x - 1) = y$,then $x = y + 1$.Substituting this into the expression:$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)}$.Performing division of $(y^2 + 2y + 1)$ by $(y - 1)$:$\frac{y^2 + 2y + 1}{y - 1} = (y + 3) + \frac{4}{y - 1}$.Thus,$\frac{1 + 2y + y^2}{y^3(y - 1)} = \frac{1}{y^3} \left( \frac{y^2 + 2y + 1}{y - 1} \right) = \frac{1}{y^3} \left( y + 3 + \frac{4}{y - 1} \right) = \frac{1}{y^2} + \frac{3}{y^3} + \frac{4}{y^3(y - 1)}$.Alternatively,using partial fraction decomposition for $\frac{x^2}{(x - 1)^3(x - 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{x - 2}$.Solving for coefficients gives $A = -4, B = -3, C = -1, D = 4$.Therefore,the expression is $\frac{-4}{x - 1} - \frac{3}{(x - 1)^2} - \frac{1}{(x - 1)^3} + \frac{4}{x - 2}$.

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

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