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