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If the denominator were $x^2+1$, the antiderivative would be arctangent. So, by completing the square, let's aim for the fraction to look like $\dfrac{1}{u^2+1}$, for some $u$. This is a good strategy for integrating an irreducible quadratic under a constant.

$$\begin{aligned} \\ \text{First: complete the square} \\ \int\frac{1}{x^2+x+1}\mathrm{d}x&= \int\frac{1}{x^2+x+\frac{1}{4}+\frac{3}{4}}\mathrm{d}x = \int\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\mathrm{d}x \\ \text{Second: get the denominator in the form $u^2+1$. To do this, we need to fix the constant} \\ &=\int\left(\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right)\left(\frac{\frac{4}{3}}{\frac{4}{3}}\right)\mathrm{d}x\\ &=\frac{4}{3}\int\frac{1}{\frac{4}{3}\cdot\left(x+\frac{1}{2}\right)^2+1}\mathrm{d}x \\ \text{Now a quick wiggle to make that first part of the denominator into something squared again:} \\ &=\frac{4}{3}\int\frac{1}{\left(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}}\right)^2+1}\mathrm{d}x \\ \text{Now we see that $u=\dfrac{2}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}$, $\mathrm{d}u=\dfrac{2}{\sqrt3}\mathrm{d}x$ will do the job} \\ &=\frac{4}{3}\int\frac{1}{u^2+1}\cdot\frac{\sqrt{3}}{2}\mathrm{d}u =\frac{2}{\sqrt3}\int\frac{1}{u^2+1}\mathrm{d}u\\ &=\frac{2}{\sqrt3}\arctan u +C\\ &=\frac{2}{\sqrt3}\arctan\left(\frac{2}{\sqrt3}x+\frac{1}{\sqrt3}\right) +C \end{aligned}$$