Somewhat Interesting

Beautiful and interesting problem!

We make a substitution $x=\frac{1}{y}$ , with $dx=-\frac{1}{y^2}dy$ , and find $\int_{1/2014}^{2014}\frac{\arctan{x}}{x}dx = \int_{1/2014}^{2014}\frac{\arctan(1/y)}{y}dy$ $= \int_{1/2014}^{2014}\frac{(\frac{\pi}{2}-\arctan{y})}{y}dy=\pi\ln{2014}-\int_{1/2014}^{2014}\frac{\arctan{x}}{x}dx$

We solve and find $\int_{1/2014}^{2014}\frac{\arctan{x}}{x}dx=\boxed{\frac{\pi}{2}\ln(2014)}$