Find $\int\frac{x+1}{x^2+x+1}dx$ Find
$\int\frac{x+1}{x^2+x+1}dx$
$\int \frac{x+1dx}{x^2+x+1}=\int \frac{x+1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
From here I don't know what to do.Write $(x+1)$ = $t$?
This does not work.Use partial integration?I don't think it will work here.
And I should complete square then find.

You will have to make the numerator as a sum of two functions like this
$\int \frac{x+\frac{1}{2}}{(x+\frac{1}{2})^2+\frac{3}{4}}dx+\int \frac{\frac{1}{2}}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
The first integral will result in $ln$ and the second in $atan$.
$\frac 1 2 \ln (x^2+x+1)+ \frac 1 {\sqrt{3}} \arctan{\frac {(2x+1)} {\sqrt{3}}}+C$
