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

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(B) $I = \int \frac{2}{1+x+x^2} dx = \int \frac{2}{(x+\frac{1}{2})^2 + \frac{3}{4}} dx$ Let $x + \frac{1}{2} = v$,then $dx = dv$. Using the formula $\

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$\frac{4}{\sqrt{3}} 	an ^{-1}\left(\frac{2 x+1}{\sqrt{3}}ight)+c$

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(B) $I = \int \frac{2}{1+x+x^2} dx = \int \frac{2}{(x+\frac{1}{2})^2 + \frac{3}{4}} dx$ Let $x + \frac{1}{2} = v$,then $dx = dv$. Using the formula $\int \frac{1}{v^2 + a^2} dv = \frac{1}{a} 	an^{-1}(\frac{v}{a}) + c$,where $a = \frac{\sqrt{3}}{2}$: $I = 2 	imes \frac{1}{\frac{\sqrt{3}}{2}} 	an^{-1}(\frac{v}{\frac{\sqrt{3}}{2}}) + c$ $I = \frac{4}{\sqrt{3}} 	an^{-1}(\frac{2v}{\sqrt{3}}) + c$ Substituting $v = x + \frac{1}{2}$: $I = \frac{4}{\sqrt{3}} 	an^{-1}(\frac{2(x + \frac{1}{2})}{\sqrt{3}}) + c = \frac{4}{\sqrt{3}} 	an^{-1}(\frac{2x+1}{\sqrt{3}}) + c$

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

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