Solveeit Logo
Login

Question

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

The partial fraction of x2(x−1)3(x−2)\frac{x^{2}}{(x - 1)^{3}(x - 2)} are

A

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

B

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

C

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

D

None of these

Answer

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

Explanation

Solution

Put the repeated factor (x−1)=y⇒x=y+1(x - 1) = y \Rightarrow x = y + 1

∴\therefore x2(x−1)3(x−2)=(1+y)2y3(y−1)=1+2y+y2y3(−1+y)\frac{x^{2}}{(x - 1)^{3}(x - 2)} = \frac{(1 + y)^{2}}{y^{3}(y - 1)} = \frac{1 + 2y + y^{2}}{y^{3}( - 1 + y)}

Dividing the numerator, 1+2y+y21 + 2y + y^{2} by −1+y- 1 + y till y3y^{3} appears as factor, we get

1+2y+y2−1+y=(−1−3y−4y2)+4y3−1+y\frac{1 + 2y + y^{2}}{- 1 + y} = ( - 1 - 3y - 4y^{2}) + \frac{4y^{3}}{- 1 + y}Given expression

= −1y3−3y2−4y+4−1+y\frac{- 1}{y^{3}} - \frac{3}{y^{2}} - \frac{4}{y} + \frac{4}{- 1 + y} = −1(x−1)3+−3(x−1)2+−4(x−1)+4(x−2)\frac{- 1}{(x - 1)^{3}} + \frac{- 3}{(x - 1)^{2}} + \frac{- 4}{(x - 1)} + \frac{4}{(x - 2)}.