TITLE: How do I evaluate $\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx$?
QUESTION [10 upvotes]: 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.

REPLY [15 votes]: What you have found is actually very useful. Let $I$ be your integral. Note that for $x>0$ one has that
$$\arctan(x)+\arctan(1/x)=\pi/2,$$
so we have the relation
$$2I=\int_{1/3}^3 \frac{\pi/2}{x^2-x+1}~dx,$$
which can be computed by completing the square on the denominator.