Use the substitution x = 1/u to find the value of
$\displaystyle \int_{\frac{1}{2}}^{2} \frac {lnx}{1+x^2}dx $
Please help me to solve this.
$\displaystyle \displaystyle x=\frac{1}{u}\Rightarrow dx=-\frac{1}{u^2}du$
$\displaystyle \displaystyle x=\frac{1}{2}\Rightarrow t=2, \ x=2\Rightarrow t=\frac{1}{2}$.
So, $\displaystyle \displaystyle I=\int_{\frac{1}{2}}^2\frac{\ln x}{1+x^2}dx=\int_2^{\frac{1}{2}}\frac{\ln u}{1+u^2}du=-\int_{\frac{1}{2}}^2\frac{\ln u}{1+u^2}du=-I\Rightarrow I=0$