int frac{x}{x^4 + x^2 + 1} dx is equal to | vedclass

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(B) Let $I = \int \frac{x}{x^4 + x^2 + 1} dx$.Substitute $u = x^2$,then $du = 2x dx$,or $x dx = \frac{1}{2} du$.The integral becomes $I = \frac{1}{2}

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{2x^2 + 1}{\sqrt{3}} ight)$

Vedclass · Answer

(B) Let $I = \int \frac{x}{x^4 + x^2 + 1} dx$.Substitute $u = x^2$,then $du = 2x dx$,or $x dx = \frac{1}{2} du$.The integral becomes $I = \frac{1}{2} \int \frac{du}{u^2 + u + 1}$.Complete the square in the denominator: $u^2 + u + 1 = (u + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$.Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{3}}{2}} 	an^{-1} \left( \frac{u + \frac{1}{2}}{\frac{\sqrt{3}}{2}} ight) + C$.$I = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{2u + 1}{\sqrt{3}} ight) + C$.Substituting $u = x^2$ back,we get $I = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{2x^2 + 1}{\sqrt{3}} ight) + C$.

$\int \frac{x}{x^4 + x^2 + 1} dx$ is equal to

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