int frac{dx}{2x^2 + x + 1} equals | vedclass

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(D) Let $I = \int \frac{dx}{2x^2 + x + 1}$.Factor out $2$ from the denominator: $I = \frac{1}{2} \int \frac{dx}{x^2 + \frac{x}{2} + \frac{1}{2}}$.Comp

Vedclass · Accepted Answer

None of these

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(D) Let $I = \int \frac{dx}{2x^2 + x + 1}$.Factor out $2$ from the denominator: $I = \frac{1}{2} \int \frac{dx}{x^2 + \frac{x}{2} + \frac{1}{2}}$.Complete the square for the quadratic expression: $x^2 + \frac{x}{2} + \frac{1}{2} = (x + \frac{1}{4})^2 - \frac{1}{16} + \frac{1}{2} = (x + \frac{1}{4})^2 + \frac{7}{16} = (x + \frac{1}{4})^2 + (\frac{\sqrt{7}}{4})^2$.Now,$I = \frac{1}{2} \int \frac{dx}{(x + \frac{1}{4})^2 + (\frac{\sqrt{7}}{4})^2}$.Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{7}}{4}} 	an^{-1} \left( \frac{x + \frac{1}{4}}{\frac{\sqrt{7}}{4}} ight) + C$.$I = \frac{1}{2} \cdot \frac{4}{\sqrt{7}} 	an^{-1} \left( \frac{4x + 1}{\sqrt{7}} ight) + C = \frac{2}{\sqrt{7}} 	an^{-1} \left( \frac{4x + 1}{\sqrt{7}} ight) + C$.Since this result does not match options $A$,$B$,or $C$,the correct option is $D$.

$\int \frac{dx}{2x^2 + x + 1}$ equals

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