<s>[INST] 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.
 [/INST] 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!
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