Indian Statistical Institute B.Math & B.Stat : Integration

Evaluate: $$\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx$$ Let $I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx$, Put $x=\frac{1}{t}$ $\implies I = -\int_{2014}^\frac{1}{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt =\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt = \int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx$ $\mathbb{Therefore,}$ $$2I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx+\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx$$ $$=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x+tan^{-1}\frac{1}{x}}{x} dx$$ $$=\int_\frac{1}{2014}^{2014} \frac{\frac{\pi}{2}}{x} dx$$ $$=\frac{\pi}{2}\int_\frac{1}{2014}^{2014} \frac{dx}{x}$$ $$=\frac{\pi}{2}[\log x]_{\frac{1}{2014}}^{2104}$$ $$=\frac{\pi}{2}\big[\log 2014-\log \frac{1}{2014}\big]$$ $$=\frac{\pi}{2}\log 2014^2$$ $$=\pi\log 2014$$