Evaluation of $\int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}x}{x} dx$ In there any way to calculate Integral


$$\int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}x}{x} dx$$


by using simple properties of definite integration. I used substitution $\tan^{-1}x=t $ and applied by parts but then I am stuck at $\int\int \csc(2t)dt$ How should I proceed? What is the most appropriate method to solve this question?
 A: For $x>0$,
$$\tan^{-1}(x)=\cot^{-1}\left(\frac1x\right)$$
Here,
$$I=\int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}x}{x} dx$$
Let $x=\frac1y$. Then,
$$\int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}x}{x} dx=\int^{\frac{1}{2014}}_{2014} \frac{\tan^{-1}\frac1y}{\frac1y}\left(-\frac1{y^2}\right)dy=-\int^{\frac{1}{2014}}_{2014} \frac{\tan^{-1}\frac1y}{y}dy=-\int^{\frac{1}{2014}}_{2014} \frac{\cot^{-1}y}{y}dy=-\int^{\frac{1}{2014}}_{2014} \frac{\frac{\pi}{2}-\tan^{-1}y}{y}dy=\int^{\frac{1}{2014}}_{2014} \frac{\tan^{-1}y-\frac{\pi}{2}}{y}dy=-I-\int^{\frac{1}{2014}}_{2014} \frac{\frac{\pi}{2}}{y}dy$$
Thus,
$$I=\int_{\frac{1}{2014}}^{2014} \frac{\pi}{4y}dy$$
which can be easily solved.
A: A straightforward one: use the substitution $u=1/x$ and the formula:
$$\forall x\in\mathbb{R}_+^*,\ \arctan\left(\frac1x\right)=\frac\pi2-\arctan(x).$$
For the sake of generality, we'll compute the integral from $1/a$ to $a$ for some $a\in\mathbb{R}_+^*$:
$$I=\int_{1/a}^{a}\frac{\arctan(x)}{x}\,\mathrm{d}x=\int_a^{1/a}u\arctan(1/u)\left(-\frac{\mathrm{d}u}{u^2}\right)=\int_{1/a}^a\frac{\pi/2-\arctan(u)}{u}\,\mathrm{d}u=\int_{1/a}^a\frac{\pi}{2u}\,\mathrm{d}u-I.$$
Hence
$$2I=\int_{1/a}^a\frac{\pi}{2u}\,\mathrm{d}u,$$
i.e.,
$$I=\frac\pi4\int_{1/a}^a\frac{\mathrm{d}u}u=\frac\pi4\bigl(\ln(a)-\ln(1/a)\bigr)=\frac\pi2\ln(a).$$
