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.
\begin{align} \int \frac{x+1}{x^2+x+1}\, dx &= \int \frac{x+\frac12 + \frac12}{x^2+x+1} \, dx \\ &= \int \frac{x+\frac12}{x^2+x+1} \, dx + \frac12 \int \frac1{x^2+x+1}\, dx \\ &=\frac12\ln |x^2+x+1| + \frac12 \int\frac1{(x+\frac12)^2 + \frac34} \, dx \end{align}
I leave the rest as an exercise.