Question

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

Easy

A

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

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

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

None of these

Solution

Put the repeated factor $(x - 1) = y \Rightarrow x = y + 1$
$∴ \frac{x^{2}}{{(x - 1)}^{3} (x - 2)} = \frac{{(1 + y)}^{2}}{y^{3} (y - 1)} = \frac{1 + 2 y + y^{2}}{y^{3} (- 1 + y)}$
Dividing the numerator,
$(1 + 2 y + y^{2}) by (- 1 + y)$ till $y^{3}$ appears as factor,
we get $\frac{1 + 2 y + y^{2}}{- 1 + y} = (- 1 - 3 y - 4 y^{2}) + \frac{4 y^{3}}{- 1 + y}$
Given expression $= \frac{- 1}{y^{3}} - \frac{3}{y^{2}} - \frac{4}{y} + \frac{4}{- 1 + y}$
$= \frac{- 1}{{(x - 1)}^{3}} + \frac{- 3}{{(x - 1)}^{2}} + \frac{- 4}{(x - 1)} + \frac{4}{(x - 2)}$

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