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

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Question

(C) To evaluate the integral $I = \int \frac{1}{x^2+x+1} \, dx$,we complete the square in the denominator: $x^2+x+1 = (x^2+x+\frac{1}{4}) + \frac{3}{4

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To evaluate the integral $I = \int \frac{1}{x^2+x+1} \, dx$,we complete the square in the denominator: $x^2+x+1 = (x^2+x+\frac{1}{4}) + \frac{3}{4} = (x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$. Substituting this into the integral: $I = \int \frac{1}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \, dx$. Using the standard integral formula $\int \frac{1}{u^2+a^2} \, du = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + c$,where $u = x+\frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$: $I = \frac{1}{\frac{\sqrt{3}}{2}} 	an^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}ight) + c$. Simplifying the expression: $I = \frac{2}{\sqrt{3}} 	an^{-1}\left(\frac{2x+1}{\sqrt{3}}ight) + c$. Thus,the correct option is $C$.

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

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