Calculate $\int \frac{1}{x^2+x+1}\mathrm{d}x$ Define the integral $I$ as follow:
$$I=\int \dfrac{1}{x^2+x+1}\mathrm{d}x.$$
I do not know how to integrate it. Any suggestions please?
I tried a lot of methods:


*

*I substituted $x^2+x=u$.

*I modified the denominator $x^2+x+1=(x+1)^2-x$ and I substituted $x+1=u$.


I know that it is somewhat very easy to do it but I am stuck.
Thanks.

complete the square as $x^2 + x + 1 = \left( x + \frac 1 2\right)^2 + \frac 3 4$, then use the substitution $\left( x + \frac 1 2\right) = u $ you will get 
$$\int \frac{1}{u^2 + \left(\frac {\sqrt 3 }{2}\right )^2}du$$
which I believe has got something to do with $\arctan(u)$
