Resolve frac{2x}{x^4 + x^2 + 1} into partial | vedclass

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

Question

(D) We have the expression $\frac{2x}{x^4 + x^2 + 1}$.First,factor the denominator: $x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1)$.N

Vedclass · Accepted Answer

$\frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1}$

Vedclass · Answer

(D) We have the expression $\frac{2x}{x^4 + x^2 + 1}$.First,factor the denominator: $x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1)$.Now,we express the fraction as partial fractions: $\frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} = \frac{Ax + B}{x^2 - x + 1} + \frac{Cx + D}{x^2 + x + 1}$.By partial fraction decomposition,we find $\frac{2x}{x^4 + x^2 + 1} = \frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1}$.Thus,the correct option is $D$.

Resolve $\frac{2x}{x^4 + x^2 + 1}$ into partial fractions.

Similar Questions

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

The values of $x$ for which $\frac{x}{(x-1)^2(x-2)}$ has a power series expansion and the coefficient of $x^n$ in such expansion are respectively:

$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A + 2B + C = ?$

If $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$,then $a = $

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