I have to fund the value of the above integral $$\int_{\frac{1}{a}}^a \frac{\arctan(x)}{x}$$ for $a=2014$.
I just saw that integral of $\frac{\arctan(x)}{x}$ does not have a closed form. So I used the substitution $x=tan(\theta)$ and then used integration by parts but did not get a result.
Let $$I(a) = \int_{\frac{1}{a}}^{a}\frac{\tan^{-1}(x)}{x}dx\;,a>0$$
Then $$I'(a) = \frac{\tan^{-1}(a)}{a}-\frac{\cot^{-1}(a)}{1}\cdot a\cdot -\frac{a}{a^2} = \frac{\tan^{-1}(a)+\cot^{-1}(a)}{a}=\frac{\pi}{2a}$$
So we get $$I(a) = \frac{\pi}{2}\ln|a|+\mathcal{C}$$
Now Put $a=1$ in $I(a)\;,$ We get $I(1) = 0$
So we get $C=0$. So $$I(a) = \frac{\pi}{2}\ln |a|$$
Since $\;a,\,1/a>0\;$ , we can assume $\;x>0\;$ , so
$$\arctan x+\arctan\frac1x=\frac\pi2\implies\text{substituting, as commented}\;\;u:=\frac1x\implies dx=-\frac{du}{u^2}$$
we get that
$$I:=\int_{1/a}^a\frac{\arctan x}xdx=\int_a^{1/a}\frac{\arctan\frac1u}{\frac1u}\left(-\frac{du}{u^2}\right)=\int_{1/a}^a\frac{\frac\pi2-\arctan u}udu$$
and now try to finish from here.
Letting $t\mapsto \frac 1t$, then $$ I=\int_{\frac{1}{x}}^x \frac{\arctan \frac{1}{t}}{t} d t $$ Averaging these two version yields $$ \frac{1}{2} \int_{\frac{1}{x}}^x \frac{\arctan t+\arctan \frac{1}{t}}{t}dt=\frac{\pi}{2} \int_{\frac{1}{x}}^x \frac{d t}{t}=\frac\pi2\ln 2 $$
