Evaluating $ \int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$ how to solve such type of definite integration?
I would like to see various methods to evaluate following integral
$$
\int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$$
 A: $$\frac12\int^\infty_{-\infty}\frac{\ln(1+16x^2)}{1+25x^2}{\rm d}x=2\pi i \operatorname*{Res}_{z=i/5}\frac{\ln(1-4iz)}{1+25z^2}=2\pi i\frac{\ln(1-4i^2/5)}{25(2i/5)}=\boxed{\Large\color{red}{{\dfrac{\pi}{5}\ln{\dfrac{9}{5}}}}}$$

Note:


*

*The imaginary part is odd and vanishes over a symmetric interval. 

*The integral over the arc vanishes as $\displaystyle \Theta\left(\frac{\ln{R}}{R}\right)$

A: Consider Following parametric Integral
$$I(\alpha )=\int_{0}^{\infty}\frac{\ln (1+\alpha ^2x^2)}{1+25x^2}\mathrm d x$$
We have $\displaystyle I(0)=0$ and $I(4)$ is required integral
By differentiating under integral sign (wrt $ \alpha $) we get
$$I'(\alpha )=\int_{0}^{\infty}\frac{2\alpha x^2}{(1+25x^2)(1+\alpha^2x^2)}\mathrm d x=\frac{\pi}{5(\pi+\alpha )}$$
Integrating back wrt $\alpha$ we get
$$I(\alpha )=\frac{1}{5} \pi  \log (\alpha +5)+c$$
Since we have $\displaystyle I(0)=0\implies c=-\frac{1}{5} \pi  \log (5)$
$$I(\alpha )=\frac{1}{5} \pi  \log (\alpha +5)-\frac{1}{5} \pi  \log (5)$$
By putting $\alpha =4$ we get required integral,
$$I(4)=\frac{1}{5} \pi  \log (9)-\frac{1}{5} \pi  \log (5)=\frac{1}{5} \pi  \log \left(\frac{9}{5}\right)$$

$$\large\int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x=\frac{1}{5} \pi  \log \left(\frac{9}{5}\right)$$

A: Rewrite the integral as
$$ 4 \log{2}\int_0^{\infty} dx \frac{1}{1+25 x^2} + \int_0^{\infty} dx \frac{\log {\left(\frac1{16}+x^2 \right)}}{1+25 x^2} $$
Consider
$$I(a) = \int_0^{\infty} dx \frac{\log{(a+x^2)}}{1+25 x^2} $$
$$\begin{align}I'(a) &= \int_0^{\infty} dx \frac{1}{(a+x^2)(1+25 x^2)}\\ &= \frac12 \int_{-\infty}^{\infty} dx \frac{1}{(a+x^2)(1+25 x^2)} \\ &= \frac1{4 \pi} \int_{-\infty}^{\infty}dk \, \frac{\pi}{\sqrt{a}} \, e^{-|k|\sqrt{a}} \frac{\pi}{5} e^{-|k|/5}\\ &= \frac{\pi}{10 \sqrt{a}} \int_0^{\infty} dk \, e^{-\left (\sqrt{a}+\frac15 \right ) k}\\ &= \frac{\pi}{10 \sqrt{a}} \frac1{\sqrt{a}+\frac15}\end{align}$$
Thus
$$I(a)= \frac{\pi}{2} \int \frac{da}{5 a + \sqrt{a} } = \pi \int \frac{dy}{5 y+1} = \frac{\pi}{5} \log{\left (\sqrt{a}+\frac15 \right )} + C$$
We find the constant of integration by evaluating
$$I(0) = 2 \int_0^{\infty} dx \frac{\log{x}}{1+25 x^2} = -\frac{\pi}{5} \log{5} \implies C=0$$
The integral is therefore
$$\frac{2 \pi \log{2}}{5}  + \frac{\pi}{5} \log{\frac{9}{20}} = \frac{\pi}{5} \log{\frac{9}{5}}$$
