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> Evaluate $$I=\int\_{1/2014}^{2014} \dfrac{\tan^{-1}x}{x}\mathrm dx$$
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$$$$I tried to solve this integral using Differentiation Under the Integral Sign. I thus redefined the integral as $$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.