TITLE: What is the value of following integral: $\int_{1/2014}^{2014}\frac{\tan^{-1}x} x \, dx$?
QUESTION [7 upvotes]: What is the value of following integral? 
  $$\int_{1/2014}^{2014}\frac{\tan^{-1}x} x \, dx$$

I am having problem evaluating this.

REPLY [2 votes]: \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 $2$.