Question: Just don't know where to start from.
Looks like this integral converges.

$\displaystyle{\int\_{0}^{\infty}\frac{\ln\left(1+x^2\right)}{1+x^2}}dx$

Answer: 
A different approach: integrate "a la Feynman". Let
$$
F(t)=\int\_{0}^{\infty}\frac{\ln(1+t^2\,x^2)}{1+x^2}\,dx.
$$
Then, differentiating inside the integral,
\begin{align\*}
F'(t)&=\int\_0^\infty\frac{2\,t\,x^2}{(1+t^2\,x^2)(1+x^2)}\,dx\\
&=\frac{2\,t}{1-t^2}\int\_0^\infty\Bigl(\frac{1}{1+t^2\,x^2}-\frac{1}{1+x^2}\Bigr)\,dx\\
&=\frac{2\,t}{1-t^2}\Bigl(\frac1t\arctan(t\,x)-\arctan x\Bigl)\Bigr|\_0^\infty\\
&=\frac{2\,t}{1-t^2}\Bigl(\frac{\pi}{2\,t}-\frac\pi2\Bigr)\\
&=\frac{\pi}{1+t}.
\end{align\*}
Integrating
$$
F(t)=\pi\log(1+t)+C.
$$
Since $\lim\_{t\to0^+}F(t)=F(0)=0$, $C=0$ and
$$
\int\_0^\infty\frac{\log(1+t^2\,x^2)}{1+x^2}\,dx=\pi\log(1+t).
$$
Of course, differentiating under the integral sign and taking the limit as $t\to0^+$ must be justified.
Yet another approach. By Euler's Beta function, for any $a>0$,
$$ f(a)=\int\_{0}^{+\infty}\frac{dx}{(1+x^2)^a} = \frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(a-\frac{1}{2}\right)}{\Gamma(a)}\tag{1}$$
hence by considering $\frac{d}{da}\log(\cdot)$ of both sides and evaluating at $a=1$, we get
$$ -\frac{f'(1)}{f(1)} = \frac{2}{\pi}\int\_{0}^{+\infty}\frac{\log(1+x^2)}{1+x^2}\,dx = \psi(1)-\psi\left(\frac{1}{2}\right)=\sum\_{n\geq 0}\left(\frac{1}{n+\frac{1}{2}}-\frac{1}{n+1}\right)\tag{2}$$
directly leading to $\int\_{0}^{+\infty}\frac{\log(1+x^2)}{1+x^2}\,dx = \color{red}{\pi\log(2)}$.