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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If $\frac{1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ and $\frac{x}{(x-1)(x-2)(x-3)}=\frac{P}{x-1}+\frac{Q}{x-2}+\frac{R}{x-3}$,then $A+2B+3C=$

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$. Then $A_r$ is equal to:

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 $y =$

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \ldots + \frac{A_n}{x+n}$. For $0 \leq r \leq n$,$A_r$ is equal to:

If $\frac{x^3+3}{(x-3)^3}=a+\frac{b}{x-3}+\frac{c}{(x-3)^2}+\frac{d}{(x-3)^3}$ then $(a+d)-(b+c)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module