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

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(D) Let $I = \int \frac{x+1}{\sqrt{x^2+x+1}} dx$. We can rewrite the numerator as $\frac{1}{2}(2x+1) + \frac{1}{2}$. So,$I = \int \frac{\frac{1}{2}(2x

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

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(D) Let $I = \int \frac{x+1}{\sqrt{x^2+x+1}} dx$. We can rewrite the numerator as $\frac{1}{2}(2x+1) + \frac{1}{2}$. So,$I = \int \frac{\frac{1}{2}(2x+1) + \frac{1}{2}}{\sqrt{x^2+x+1}} dx = \frac{1}{2} \int \frac{2x+1}{\sqrt{x^2+x+1}} dx + \frac{1}{2} \int \frac{1}{\sqrt{x^2+x+1}} dx$. Let $I_1 = \frac{1}{2} \int \frac{2x+1}{\sqrt{x^2+x+1}} dx$. Substituting $u = x^2+x+1$,$du = (2x+1)dx$,we get $I_1 = \frac{1}{2} \int u^{-1/2} du = \frac{1}{2} (2\sqrt{u}) = \sqrt{x^2+x+1}$. Let $I_2 = \frac{1}{2} \int \frac{1}{\sqrt{(x+1/2)^2 + (\sqrt{3}/2)^2}} dx$. Using the formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \sinh^{-1}(\frac{x}{a})$,we get $I_2 = \frac{1}{2} \sinh^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}\right) = \frac{1}{2} \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)$. Thus,$I = \sqrt{x^2+x+1} + \frac{1}{2} \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C$.

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

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