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I've to evaluate the integral $\int_1^4 \frac{dx}{x^2+x+1}$

but I can't find the answer. I checked with Wolfram Alpha but I still don't fully understand. Could you please explain the steps to me? I think I should use arctan in my answer.

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    This is *not* an indefinite integral, but I don't have tag privileges yet... could someone help me out? :)2012-11-21

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First note that $\dfrac1{x^2 + x + 1} = \dfrac1{(x+1/2)^2 + (\sqrt{3}/2)^2}$ We now have that \begin{align} I = \int_1^4 \dfrac{dx}{x^2 + x + 1} = \int_1^4 \dfrac{dx}{(x+1/2)^2 + (\sqrt{3}/2)^2} \end{align} Now let $y = x+1/2$, then we get that $I = \int_{3/2}^{9/2} \dfrac{dy}{y^2 + (\sqrt{3}/2)^2}$ Now let $y = \sqrt{3}/2 \tan(\theta)$. We then get that $dy = \sqrt{3}/2 \sec^2(\theta) d \theta$. Hence, $I = \int_{\theta = \arctan(\sqrt{3})}^{\theta = \arctan(3 \sqrt{3})} \dfrac{\sqrt{3}/2 \sec^2(\theta) d \theta}{3/4 \sec^2(\theta)} = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \arctan(\sqrt{3}) \right) = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \dfrac{\pi}3 \right)$

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    I do know that, cause that is in my book. But ok then I need to study some more integrals :)2012-11-21