$$\int\frac{x+1}{x^2+x+1}dx=\int\frac{x+1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}dx$$

Letting $t=x+\frac{1}{2}$ yields

$$\begin{align*}\int\frac{t+\frac{1}{2}}{t^2+\frac{3}{4}}dt&=\int\frac{t}{t^2+\frac{3}{4}}dt+\frac{1}{2}\int\frac{1}{t^2+\frac{3}{4}}dt\\&=\frac{1}{2}\int\frac{d\left(t^2+\frac{3}{4}\right)}{t^2+\frac{3}{4}}+\frac{\sqrt{3}}{3}\int\frac{d\left(\frac{2}{\sqrt{3}}t\right)}{1+\left(\frac{2}{\sqrt{3}}t\right)^2}\\&=\frac{1}{2}\ln\left(t+\frac{3}{4}\right)+\frac{\sqrt{3}}{3}\arctan\left(\frac{2}{\sqrt{3}}t\right) + C\\&=\boxed{\frac{1}{2}\ln(x^2+x+1)+\frac{\sqrt{3}}{3}\arctan\frac{2x+1}{\sqrt{3}}+C}\end{align*}$$