If frac{x}{(x-1)(x^2+1)^2} = frac{1}{4}left[f | vedclass

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

Question

(A) We use the method of partial fractions for the expression $\frac{x}{(x-1)(x^2+1)^2}$.Let $\frac{x}{(x-1)(x^2+1)^2} = \frac{A}{x-1} + \frac{Bx+C}{x

Vedclass · Accepted Answer

$\frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$

Vedclass · Answer

(A) We use the method of partial fractions for the expression $\frac{x}{(x-1)(x^2+1)^2}$.Let $\frac{x}{(x-1)(x^2+1)^2} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}$.Multiplying by $(x-1)(x^2+1)^2$,we get $x = A(x^2+1)^2 + (Bx+C)(x-1)(x^2+1) + (Dx+E)(x-1)$.Setting $x=1$,we get $1 = A(2)^2$,so $A = \frac{1}{4}$.Comparing coefficients or substituting values for $x$,we find $B = -\frac{1}{4}$,$C = -\frac{1}{4}$,$D = -\frac{1}{2}$,and $E = \frac{1}{2}$.Substituting these values,we have $\frac{x}{(x-1)(x^2+1)^2} = \frac{1}{4(x-1)} - \frac{x+1}{4(x^2+1)} + \frac{-x/2 + 1/2}{(x^2+1)^2}$.This simplifies to $\frac{1}{4}\left[\frac{1}{x-1} - \frac{x+1}{x^2+1}\right] + \frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$.Comparing this with the given equation,we get $y = \frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$.

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 $y =$

Similar Questions

If $\frac{x^4+x^3+2x^2-2x+1}{x^3+x^2} = P(x) + \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$,then $A+B+C = $

If $\frac{x^4}{(x-1)(x-2)(x-3)}=Ax+B \cdot \frac{1}{x-1}+C \cdot \frac{1}{x-2}+D \cdot \frac{1}{x-3}+E$,then $A+B+C+D+E=$

If $\frac{(x - 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then

Let $H(x)=3x^4+6x^3-2x^2+1$ and $g(x)$ be a polynomial of degree one. If $\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$, then $H(-1)+2H(2)-3H(1)=$

If $\frac{3x^2+x+1}{(x-1)^4} = \frac{a}{(x-1)} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4}$,then $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module