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

Question

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.

Answer 1

\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.

Answer 2

Hint: Write $(x+1)=(2x+1)/2+1/2$

Answer 3

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 $atan$.

$\frac 1 2 \ln (x^2+x+1)+ \frac 1 {\sqrt{3}} \arctan{\frac {(2x+1)} {\sqrt{3}}}+C$

Answer 4

Write $x+\frac{1}{2} = t.$ Then,

$$\frac{x+1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} = \frac{x+\frac12 + \frac12}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} = \frac{t}{t^2+\frac{3}{4}} + \frac{\frac12}{t^2+\frac{3}{4}}.$$

Can you integrate these two terms now?