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.

$$I := \int^1_0 \frac{dx}{x^2+x+1}$$
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
$$I = \frac{\pi}{3\sqrt{3}} \approx .605$$
