TITLE: Evaluating $ \int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$
QUESTION [7 upvotes]: how to solve such type of definite integration?
I would like to see various methods to evaluate following integral
$$
\int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$$

REPLY [6 votes]: $$\frac12\int^\infty_{-\infty}\frac{\ln(1+16x^2)}{1+25x^2}{\rm d}x=2\pi i \operatorname*{Res}_{z=i/5}\frac{\ln(1-4iz)}{1+25z^2}=2\pi i\frac{\ln(1-4i^2/5)}{25(2i/5)}=\boxed{\Large\color{red}{{\dfrac{\pi}{5}\ln{\dfrac{9}{5}}}}}$$

Note:

The imaginary part is odd and vanishes over a symmetric interval. 
The integral over the arc vanishes as $\displaystyle \Theta\left(\frac{\ln{R}}{R}\right)$