# Evaluate $\int_1^4\frac{dx}{x^2+x+1}$

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? :) – apnorton Nov 21 '12 at 0:55

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, $$I = \int_{\theta = \arctan(\sqrt{3})}^{\theta = \arctan(3 \sqrt{3})} \dfrac{\sqrt{3}/2 \sec^2(\theta) d \theta}{3/4 \sec^2(\theta)} = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \arctan(\sqrt{3}) \right) = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \dfrac{\pi}3 \right)$$
Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}u$. You will end up (apart from constant in front) integrating $\frac{1}{u^2+1}\,du$. – André Nicolas Nov 21 '12 at 0:53
Oh, yeah I guess that's "easier". I first substituted $u=x+.5$ and afterwards $v=\frac{2u}{\sqrt3}$. But it appears to be quite hard... These are questions I have to solve "at home" (so that means, the easy ones) and regarding that I only had 1 hour lecture about integrals I don't think I should be capable of doing this.... Do you think I should? – Bob Nov 21 '12 at 0:57
@Bob If you are familiar with the fact that $$\int \dfrac{du}{u^2+1} = \arctan(u) + c$$ then I do not think that this is hard and you should be able to do it. – user17762 Nov 21 '12 at 0:59