Đặt $x=\dfrac{1}{t} \Rightarrow dx =-\dfrac{1}{t^2}dt$
$I=-\int \limits_{2}^{\frac{1}{2}} \dfrac{\ln \dfrac{1}{t}}{t^2 (\dfrac{1}{t^2}+1)}dt=\int \limits_{\frac{1}{2}}^2 \dfrac{\ln t^{-1}}{t^2+1}dt$
$=-\int \limits_{\frac{1}{2}}^2 \dfrac{\ln t}{t^2+1}dt =-\int \limits_{\frac{1}{2}}^2 \dfrac{\ln x}{x^2+1}dx =-I$
$\Rightarrow 2I = 0 \Rightarrow I=0$