[tex]u=\frac{1}{x}\Rightarrow \int{\dfrac{\mathrm{tg}\left(\frac{1}{x}\right)}{{x}^{2}}}{\;\mathrm{d}x}=\int{-\dfrac{\sin\left(u\right)}{\cos\left(u\right)}}{\;\mathrm{d}u}\overset{v=\cos(u)}{=}\int{\dfrac{1}{v}}{\;\mathrm{d}v}=\\=\ln\left(\left|v\right|\right)+C=\ln\left(\left|\cos\left(u\right)\right|\right)+C=\ln\left(\left|\cos\left(\frac{1}{x}\right)\right|\right)+C[/tex]