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

$\frac{x^4}{x^3-3x+2}$ is a

If $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$ and $a>b$ then $B+D=$

If $\frac{x^2+x+1}{x^2+2x+1}=A+\frac{B}{x+1}+\frac{C}{(x+1)^2}$,then $A-B$ is equal to

If $\frac{3x+1}{(x-1)(x^2+2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2}$,then $5(A-B)=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module