How do I evaluate $\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx$? I need to calculate the following definite integral:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx.$$
The only thing that I've found is:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx = \int_{1/3}^3 \frac{\arctan \frac{1}{x}}{x^2 - x + 1} \; dx,$$
but it doesn't seem useful.

Hint
$$\arctan{x}+\arctan{\dfrac{1}{x}}=\dfrac{\pi}{2}$$
so
$$\int_{\frac{1}{3}}^{3}\dfrac{\arctan{x}}{x^2-x+1}dx=I$$
let
$x=\dfrac{1}{u}$,then
$$I=\int_{\frac{1}{3}}^{3}\dfrac{\arctan{\frac{1}{x}}}{x^2-x+1}dx$$
so
$$2I=\dfrac{\pi}{2}\cdot\int_{\frac{1}{3}}^{3}\dfrac{1}{x^2-x+1}dx$$
