<s>[INST] 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.
 [/INST] 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,
$$I = \int_{\theta = \arctan(\sqrt{3})}^{\theta = \arctan(3 \sqrt{3})} \dfrac{\sqrt{3}/2 \sec^2(\theta) d \theta}{3/4 \sec^2(\theta)} = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \arctan(\sqrt{3}) \right) = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \dfrac{\pi}3 \right)$$
</s>