Integrate $\int \frac{1}{x^2+x+1}dx$

First note that
$$x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=(x+\frac{1}{2})^2+\frac{3}{4}.$$
So
\[\int \frac{1}{1+x+x^2}dx=\int \frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx\]
letting $u=x+\frac{1}{2}$, we get $du=dx$ and hence
\[= \int \frac{1}{u^2+\frac{3}{4}}du = \int \frac{1}{u^2+(\frac{\sqrt{3}}{2})^2}du\]
\[=\frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}( \frac{u}{\frac{\sqrt{3}}{2}})+c=\frac{2}{\sqrt{3}}\tan^{-1}(\frac{2}{\sqrt{3}}u)+c= \frac{2}{\sqrt{3}}\tan^{-1}(\frac{2}{\sqrt{3}}(x+\frac{1}{2}))+c.\]
We have used the integral formula 
\[\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\tan^{-1} \frac{x}{a}+c.\]