Evaluate $\int_{1/2014}^{2014} \frac{\tan^{-1}x}{x}dx$ using Differentiation Under the Integral Sign > Evaluate $$I=\int_{1/2014}^{2014} \dfrac{\tan^{-1}x}{x}\mathrm dx$$ $$$$I tried to solve this integral using Differentiation Under the Integral Sign. I thus redefined the integral as $$I(a)=\int_{1/2014}^{2014} \frac{\tan^{-1}(ax)}{x}\mathrm dx$$$$\Rightarrow I'(a)= \int_{1/2014}^{2014} \frac{1}{1+a^2x^2}\mathrm dx=\left(\dfrac{\tan^{-1}(ax)}{a}\right)_{1/2014}^{2014}$$ I know that there are other methods of solving this integral. However for the sake of practice, I am specifically interested in a solution involving Differentiation Under the Integral Sign.

Note that \begin{eqnarray*} \tan^{-1}(x)+\tan^{-1}(1/x)=\frac{\pi}{2}. \end{eqnarray*} and \begin{eqnarray*} I= \int_{1/2014}^{2014} \frac{\tan^{-1}(x)}{x} dx =\frac{\pi}{2} \int_{1/2014}^{2014} \frac{dx}{x} + \int_{1/2014}^{2014} \frac{\tan^{-1}(1/x)}{1/x} \frac{-dx}{x^2}. \end{eqnarray*} Now substitute $u=1/x$ in the latter integral to obtain $-I$. ... should be a doddle from here ?