Mad Integral

Let $F(a)= 2\displaystyle\int_0^\infty \dfrac {\ln(1+a^2x^2)}{1+x^2} \mathrm{d}x~~,\small{F(0)=0}$

$F'(a)= 4a\int_0^\infty \dfrac {x^2}{(1+a^2x^2)(1+x^2)} \mathrm{d}x$

$=\dfrac{4a}{a^2-1}\int_0^\infty \left[\dfrac {1}{1+x^2}-\dfrac{1}{1+a^2x^2}\right] \mathrm{d}x$

$=\dfrac{4a}{a^2-1}\left[\tan^{-1} x-\dfrac 1a\tan^{-1} ax\right]_0^{\infty}$

$=\dfrac{4a}{(a-1)(a+1)}\left[\dfrac{\pi}2-\dfrac{\pi}{2a}\right]$

$=\dfrac{2\pi}{a+1}\implies F(a)=\int\dfrac{2\pi\mathrm{d}a}{a+1}$

$\implies F(a)=2\pi\ln(a+1)~~~~\small{(F(0)=0)}$

In question we need to evaluate $F(1)$ which comes out to be $2\pi\ln(1+1)=\pi \ln \boxed{\color{#D61F06}{4}}$ .

$\begin{aligned} I & = \int_{-\infty}^\infty \frac {\ln(x^2+1)}{x^2+1} dx & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = 2 \int_0^\infty \frac {\ln(x^2+1)}{x^2+1} dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = 2 \int_0^\frac \pi 2 \ln(\sec^2 \theta)\ d\theta \\ & = \color{#D61F06} 4 \int^0_\frac \pi 2 \ln(\cos \theta) \ d\theta & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = 2 \int^0_\frac \pi 2 (\ln(\cos \theta) + {\color{#D61F06}\ln(\sin \theta)}) \ d\theta \\ & = 2 \int^0_\frac \pi 2 \ln \left(\frac {\sin 2 \theta}2\right) d\theta \\ & = {\color{#3D99F6}2 \int^0_\frac \pi 2 \ln (\sin 2 \theta)\ d\theta} + 2 \int_0^\frac \pi 2 \ln 2\ d\theta & \small \color{#3D99F6} \text{Let }\phi = 2 \theta \implies d\phi = 2 \ d\theta \\ & = {\color{#3D99F6}\int^0_\pi \ln (\sin \phi)\ d\phi} + 2 \int_0^\frac \pi 2 \ln 2\ d\theta & \small \color{#3D99F6} \text{Since }\sin \phi \text{ is symmetrical about }\frac \pi 2 \\ & = {\color{#D61F06}2 \int^0_\frac \pi 2 \ln (\sin \phi)\ d\phi} + \pi \ln 2 & \small \color{#D61F06} \text{Note that } \int^0_\frac \pi 2 \ln (\sin \theta)\ d\theta = \int^0_\frac \pi 2 \ln (\cos \theta)\ d\theta = \frac I4 \\ & = {\color{#D61F06}\frac I2} + \pi \ln 2 \\ & = 2\pi \ln 2 = \pi \ln 4 \end{aligned}$

Therefore, $a=\boxed{4}$ .