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.
Changing variable $x\to\tan x$, one has
$$ \int\_{0}^{\infty}\frac{\ln\left(1+x^2\right)}{1+x^2}dx=\int\_{0}^{\frac\pi2}\ln(\sec^2x)dx=\frac{\pi}{2}\ln2+\int\_{0}^{\frac\pi2}\ln(\sec x)dx=\pi\ln2.$$
Here
$$ \int\_{0}^{\frac\pi2}\ln(\sec x)dx=\frac\pi2\ln2. $$