Evaluate $\int_{1/2014}^{2014} \frac{\tan^{-1}x}{x}dx$ using Differentiation Under the Integral Sign

This is too long for a comment; however, this might shed some light on why differentiation under the integral sign leads back to the same integral: $$ \begin{align} &\int_0^1\left(\color{#C00}{\frac{\mathrm{d}}{\mathrm{d}a}\int_{1/2014}^{2014}\frac{\tan^{-1}(ax)}x\,\mathrm{d}x}\right)\mathrm{d}a\tag1\\ &=\int_0^1\left(\frac{\mathrm{d}}{\mathrm{d}a}\int_{a/2014}^{2014a}\frac{\tan^{-1}(x)}x\,\mathrm{d}x\right)\mathrm{d}a\tag2\\ &=\int_0^1\left(\color{#C00}{\frac{\tan^{-1}(2014a)}a-\frac{\tan^{-1}\left(\frac{a}{2014}\right)}a}\right)\mathrm{d}a\tag3\\ &=\int_0^{2014}\frac{\tan^{-1}(a)}a\,\mathrm{d}a-\int_0^{1/2014}\frac{\tan^{-1}(a)}a\,\mathrm{d}a\tag4\\ &=\int_{1/2014}^{2014}\frac{\tan^{-1}(a)}a\,\mathrm{d}a\tag5 \end{align} $$ Explanation:
$(1)$: the integral for $a=0$ is $0$
$(2)$: substitute $x\mapsto x/a$
$(3)$: Fundamental Theorem of Calculus
$(4)$: substitute $a\mapsto a/2014$ on the left and $a\mapsto2014a$ on the right
$(5)$: subtract the integrals

No matter how you slice it, the red expressions are the same, and you will run back into the original integral.

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 ?

Thus $$I' (a) =\int_{2014^{-1}}^{2014} \frac{1}{1+a^2 x^2 } dx =a^{-1}\int_{2014^{-1}a}^{2014 a}\frac{1}{1+t^2 } dt=a^{-1}\arctan t|_{2014^{-1}a}^{2014 a}$$