How to take $\int^1_{-1} \frac{xdx}{x^2+x+1}$? Here's the integral:
$$\int^1_{-1} \frac{xdx}{x^2+x+1}$$
So I am basicly confused about $xdx$ if it were only $dx$ I would just compleate the square and go ahead with table formula, but as i can judge it doesn't work properly witx $xdx$
Can someone give me a hint, please?

Let $I = \int^1_{-1} \frac{x}{x^2+x+1} dx$
$2I = \int^1_{-1} \frac{2x + 1 - 1}{x^2+x+1} dx$
$2I = [\ln|x^{2}+x+1|]_{-1}^{1} -\int_{-1}^{1}\frac{1}{x^{2}+x+1}dx$
$2I = \ln3 -\int_{-1}^{1}\frac{1}{(x+\frac{1}{2})^{2}+\frac{3}{4}}dx$
Can you solve it from there?
Edit: In case you get stuck:
Let $x+\frac{1}{2} = \frac{\sqrt{3}}{2}\tan u$
$\Rightarrow dx = \frac{\sqrt{3}}{2}\sec^{2}u$
Lower limit: 
$-1+\frac{1}{2} = \frac{\sqrt{3}}{2}\tan u\Rightarrow \tan u = -\frac{1}{\sqrt{3}} \Rightarrow u = -\frac{\pi}{3}$
Upper limit:
$\frac{3}{2}=\frac{\sqrt{3}}{2}\tan u \Rightarrow u = \frac{\pi}{6}$ 
$2I = \ln3 - \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{\frac{3}{4}(\tan^{2}u+1)} \cdot \frac{\sqrt{3}}{2}\sec^{2}u\ du = \ln3 - \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{2}{\sqrt{3}}\ du$
$2I = \ln 3 - (\frac{2\pi}{3\sqrt{3}} +\frac{2\pi}{6\sqrt{3}}) = \ln 3 -\frac{\pi}{\sqrt{3}}$
$I = \frac{1}{2}(\ln 3 - \frac{\pi}{\sqrt{3}})$
