Let,

$I={\int }_{\frac{1}{2}}^2\frac{{\tan }^{-1}x}{x}dx ........\left(1\right)$

Now, let $x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$

So,

$I=-{\int }_2^{\frac{1}{2}}\frac{{\tan }^{-1}\left({\displaystyle \frac{1}{t}}\right)}{{\displaystyle \frac{1}{t}}}\times \frac{1}{t^2}dt$

$\Rightarrow I=-{\int }_2^{\frac{1}{2}}{\tan }^{-1}\left(\frac{1}{t}\right)\times \frac{1}{t}dt$

$\Rightarrow I={\int }_{\frac{1}{2}}^2{\cot }^{-1}t\times \frac{1}{t}dt$

$\Rightarrow I={\int }_{\frac{1}{2}}^2{\cot }^{-1}x\times \frac{1}{x}dx .......\left(2\right)$

Now adding $\left(1\right) \& \left(2\right)$, we get,

$2I={\int }_{\frac{1}{2}}^2\frac{1}{x}\left({\cot }^{-1}x+{\tan }^{-1}x\right)dx$

$\Rightarrow 2I={\int }_{\frac{1}{2}}^2\frac{1}{x}\times \frac{\mathrm{\pi }}{2}dx$

$\Rightarrow 2I=\frac{\mathrm{\pi }}{2}{\left[\ln x\right]}_{\frac{1}{2}}^2$

$\Rightarrow 2I=\frac{\mathrm{\pi }}{2}\left[{\log }_{e}2-{\log }_e\frac{1}{2}\right]$

$\Rightarrow 2I=\frac{\mathrm{\pi }}{2}\left[2{\log }_{e}2\right]$

$\Rightarrow I=\frac{\mathrm{\pi }}{2}{\log }_{e}2$