Evaluate $\int^1_0 \frac{dx}{x^2+x+1}$ Here is an integral which I don't know the answer:
$$\int^1_0 \frac{dx}{x^2+x+1}$$
I tried to use complex number to solve i.e. the root of $x^2+x+1$ is $(-1/2 +  \sqrt {3}i/2)$.
Let w=$(-1/2 +  \sqrt {3}i/2)$ , then it becomes $ \int_{0}^{1} 1/(x-w)^2\,\mathrm{d} (x-w)$ , the answer is in terms of complex number. I wanna ask whether my method is correct and seek for another method. Thank you.
 A: $$I := \int^1_0 \frac{dx}{x^2+x+1}$$
This looks like a job for . . . $\arctan$! 
Recall that 
$$\int\frac{du}{u^2+a^2} = \frac { \arctan \left( \frac  {u}{a} \right)} {a} + C$$
Complete the square in the bottom to get
$$I = \int^1_0 \frac{dx} {{\left(x+\frac{1}{2} \right)}^2 + \frac{3}{4}}$$
Let $u = x + \frac{1}{2}, du = dx$.
$$ \frac { 2\arctan \left( \frac{2u}{\sqrt{3}} \right) } {\sqrt{3}}$$
Back-substitute $u=x+\frac{1}{2}$ to get:
$$ \frac { 2\arctan \left( \frac{2x+1}{\sqrt{3}}\right) } {\sqrt{3}}$$
Now just evaluate at your endpoints to get
$$I = \frac{\pi}{3\sqrt{3}} \approx .605$$
A: Your idea of using complex functions can work, but you seem to believe the polynomial has only one root whereas it actually has two different (complex conjugate, of course, as the polynomial is real) roots:
$$x^2+x+1=\left(x+\frac{1+\sqrt 3i}{2}\right)\left(x+\frac{1-\sqrt 3i}{2}\right)=(x+w)(x+\overline w)\,\,,\,w:=\frac{1+\sqrt 3i}{2}$$
(note $\,w\,$ is a root of unit of order 6, though this is important here only to write $\,|w| = 1\,$) and from here we get
$$\frac{1}{x^2+x+1}=\frac{1}{(x+w)(x+\overline w)}=\frac{1}{\sqrt 3\,i}\left(\frac{1}{x+\overline w}-\frac{1}{x+w}\right)\Longrightarrow$$
$$\Longrightarrow \int_0^1\frac{dx}{x^2+x+1}=\frac{1}{\sqrt 3\,i}\left(\int_0^1\frac{dx}{x+\overline w}-\int_0^1\frac{dx}{x+w}\right)=\frac{1}{\sqrt 3\,i}\left.\left(Log(x+\overline w)-Log(x+w)\right)\right|_0^1$$
Now choose the principal branch of the complex logarithm, and we get:
$$\int_0^1\frac{dx}{x^2+x+1}=\frac{1}{\sqrt 3\,i}\left.Log\left(\frac{x+\overline w}{x+w}\right)\right|_0^1=\frac{1}{\sqrt 3\,i}\left.Log\left(\frac{(x+\overline w)^2}{|x+ w|^2}\right)\right|_0^1=$$
$$=\frac{1}{\sqrt 3\,i}\left(Log\frac{(1+\overline w)^2}{|1+w|^2}-Log\frac{\overline w^2}{|w|^2}\right)=\frac{1}{\sqrt 3\,i}\left[Log\left(\frac{1}{2}-\frac{\sqrt 3}{2}i\right)-Log\left(-\frac{1}{2}-\frac{\sqrt 3}{2}i\right)\right]=$$
$$=\frac{1}{\sqrt 3\,i}\,i\left[\frac{5\pi}{3}-\frac{4\pi}{3}=\right]=\frac{\pi}{3\sqrt 3}$$
You get, of course, the same result as the first, much simpler, answer by Joe, and there you don't have to mess with complex functions, their branches, conjugates and other beasts...yet it is doable!
A: Note that 
$$x^2+x+1=\frac{1}{4}(4x^2+4x+4)=\frac{1}{4}((2x+1)^2+3).$$
Make the change of variable $2x+1=\sqrt{3}u$. When the smoke clears, you will have a very familiar integral.
Remark: The complex number approach can be made to work. The details in the post are not correct, the expression at the bottom has two distinct roots. Once that part is cleaned up, one can use partial fractions. The answer is in terms of complex logarithms, which can be treacherous. Not recommended unless one has full control of the situation!  But it brings out an interesting connection between the complex logarithm and the $\arctan$ function.
A: You can complete the square with the denominator.  Think how you would turn it into an arctan integral.
$$\int \frac{1}{x^2+1}\ \mathrm{d}x=\tan^{-1}x+C$$
A: \begin{eqnarray}
\int_{0}^{1} \dfrac{1}{x^2+x+1} dx &=& \int_{0}^{1} \dfrac{1}{(x+1/2)^2 +3/4} dx \\
&=& \int_{1/2}^{3/2} \dfrac{1}{y^2 +3/4}dy \\
&=& \dfrac{4}{3}\int_{1/2}^{3/2} \dfrac{1}{(\dfrac{2}{\sqrt{3}}y)^2 + 1} dy \\
&=& \dfrac{\sqrt{3}}{2} \dfrac{4}{3} \int_{\sqrt{3}/3}^{\sqrt{3}} \dfrac{1}{z^2+1} dz\\
&=& 2 \dfrac{\sqrt{3}}{3} (\arctan(\sqrt{3}) - \arctan(\sqrt{3}/3))\\
&=&2 \dfrac{\sqrt{3}}{3}( \dfrac{\pi}{3} - \dfrac{\pi}{6})\\
&=& \dfrac{\sqrt{3}\pi}{9}\\
&=& 0,6045997880781
\end{eqnarray}
