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

I am having problem evaluating this.
 A: 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 $x>0$.
A: \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 $2$.
A: 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
$$I = \frac{\pi}{2}\ln(2014)$$
