Have I Seen This Before?

Usually when we see $x^2+1$ the sub that comes to mind is $x=\tan t$ like What @Ali Caglayan has done. Here's another (and more general) solution :

For $a\in \mathbb{R}$ set : $I(a) = \int_0^{\infty} \frac{\log(1+a^2 x^2)} {1+x^2}\ \mathrm{d}x$ Then $\begin{aligned}I'(a)&= \frac{2a}{a^2-1} \int_0^{\infty}\frac{1}{x^2+1}- \frac{1}{a^2 x^2+1 } \ \mathrm{d}x\\ & = \frac{2a}{a^2-1} \left(\frac{\pi}{2}- \frac{\pi}{2a} \right)\\ &=\frac{\pi}{1+a} \end{aligned}$ Since $I(0)=0$ we conclude that : $I(a)= \int_0^a \frac{\pi}{1+t} \ \mathrm{d}t =\pi\ln(1+a)$ Just set $a=1$ to get the result that we want.

This is a beautiful problem! First start of with $I = \int_0^\infty\frac{\log(x^2+1)}{x^2+1}\mathrm dx$ We substitute $\tan \theta = x$ giving us $\sec^2\theta\ \mathrm d\theta=\mathrm dx$ and after some cancelling using $\sec^2x=\tan^2x+1$ we get $I=\int_0^{\pi/2}\log(\sec^2 \theta)\ \mathrm d\theta$ Which can be rewritten as $I=-2\int_0^{\pi/2}\log(\cos \theta)\ \mathrm d\theta$ If we map $\theta \mapsto \frac \pi2-\theta$ we see by symmetry $I=-2\int_0^{\pi/2}\log(\sin \theta)\ \mathrm d\theta$ Now we map $\theta \mapsto 2\theta$ and use symmetry of the sin curve to split the integral up again $I = -\int_0^\pi \log(\sin 2\theta)\ \mathrm d\theta=-2\int_0^{\pi/2}\log(\sin 2\theta)\ \mathrm d\theta$ Now $\sin 2\theta = 2\sin \theta \cos \theta$ , so $\begin{aligned} I &= 2 \int_0^{\pi/2}\log(2\sin \theta\cos \theta)\ \mathrm d\theta\\ I &=-\pi \log 2 -2\int_0^{\pi/2}\log(\sin\theta)\ \mathrm d\theta-2\int_0^{\pi/2}\log(\cos \theta)\ \mathrm d\theta\\ \implies I &= -\pi \log 2 + 2I \end{aligned}$ So by symmetry of the integrand we rearrange the equation to get $I=\pi \log 2\approx \boxed{2.1776}$ What a beautiful result!