Use the substitution $x=\frac{1}{u}$ to find the value of
$\int^2_{\frac{1}{2}} \, \frac{ln x}{1+x^2} dx$
2. $\frac{1}{z}=x\Rightarrow - \frac{dz}{z^2}=dx$ then $I=\int_\frac{1}{2}^2\frac{\log x}{1+x^2}dx=\int_2^\frac{1}{2}\frac{-\log\frac{1}{z}}{1+\frac{1}{z^2}}\frac{1}{z^2}dz=-\int_\frac{1}{2}^2\frac{\log z}{1+z^2}dz=-I$ $\Rightarrow I=-I\Rightarrow I=0.$