What is the value of following integral: $\int_{1/2014}^{2014}\frac{\tan^{-1}x} x \, dx$?

Let $x = \frac{1}{u}, dx = -\frac{1}{u^2} du$. Then $$I=\int_{1/2014}^{2014} \frac{\tan^{-1}(x)}{x}dx = \int_{2014}^{1/2014} -\frac{\tan^{-1}(1/u)}{u}du$$ For positive $u$, we have $\tan^{-1}(1/u) = \frac{\pi}{2}-\tan^{-1}(u)$, and so $$= \int_{1/2014}^{2014} \frac{\pi}{2u} - \frac{\tan^{-1}(u)}{u}\, du = \int_{1/2014}^{2014} \frac{\pi}{2u}\, du -I$$ and so $$2I = \frac{\pi}{2}\int_{1/2014}^{2014} \frac{1}{u}\, du = \pi \ln(2014)$$ giving us $$I = \frac{\pi}{2}\ln(2014)$$

I suggest to try the substitution $y=1/x$ and use the fact that $\tan^{-1}x+\tan^{-1}(1/x)=\dfrac{\pi}{2}$ for $x>0$.

\begin{align} w & = \frac 1 x \\[10pt] dw & = \frac{-dx}{x^2}, \text{ so } \frac{dw} w = \frac{-dx} x. \\[20pt] \int_{1/2014}^{2014} \frac{\arctan x} x \, dx & = \int_{2014}^{1/2014} \frac{\arctan(1/w)}{w} (-dw) \\[10pt] & = \int_{2014}^{1/2014} \frac{\frac \pi 2 - \arctan w} w \, (-dw) \\[10pt] & = \int_{1/2014}^{2014} \frac \pi {2w} \, dw - \int_{1/2014}^{2014} \frac {\arctan w} w \, dw. \\[10pt] \text{So } I & = \int_{1/2014}^{2014} \frac \pi {2w} \, dw - I, \\[10pt] \text{and thus } 2I & = \int_{1/2014}^{2014} \frac \pi {2w}\, dw, \end{align} and then divide both sides by $2$.