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

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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.

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