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 $\frac{x-4}{x^2-5x+6}$ can be expanded in the ascending powers of $x$,then the coefficient of $x^3$ is

$\begin{aligned} & \text{If } \frac{x^3}{(2x-1)(x-1)^2} = A + \frac{B}{2x-1} + \frac{C}{x-1} \\ & + \frac{D}{(x-1)^2}, \text{ then } 2A - 3B + 4C + 5D = \end{aligned}$

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

If $\frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + \frac{q}{2x - 1} + \frac{r}{x + 2} + \frac{s}{x - 3}$,then which of the following is true?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module