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.
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\begin{align}
\int\_{0}^{\infty}{\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x & =
-2\int\_{0}^{\pi/2}\ln\pars{\cos\pars{\theta}}\,\dd\theta =
\left.-2\,\Re\int\_{0}^{\pi/2}\ln\pars{z + 1/z \over 2}\,{\dd z \over \ic z}\,
\right\vert\_{\ z\ =\ \exp\pars{\ic\theta}}
\\[5mm] & =
\left.-2\,\Im\int\_{0}^{\pi/2}\ln\pars{1 + z^{2} \over 2z}\,{\dd z \over z}\,
\right\vert\_{\ z\ =\ \exp\pars{\ic\theta}}
\\[1cm] & \stackrel{\mrm{\epsilon}\ \to\ 0^{+}}{\sim}
2\,\Im\int\_{1}^{\epsilon}\bracks{%
\ln\pars{1 - y^{2} \over 2y} - {\pi \over 2}\,\ic}\,{\dd y \over y}
\\ & +
2\,\Im\int\_{\pi/2}^{0}\bracks{%
-\ln\pars{2} - \ln\pars{\epsilon} - \theta\,\ic}\ic\,\dd\theta\ +\
\overbrace{2\,\Im\int\_{\epsilon}^{1}\ln\pars{1 + x^{2} \over 2x}
\,{\dd x \over x}}^{\ds{=\ 0}}
\\[1cm] & =
-\pi\ln\pars{\epsilon} + 2\bracks{-\ln\pars{2} - \ln\pars{\epsilon}}\pars{-\,{\pi \over 2}} \,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{=}\,\,\,
\bbx{\ds{\pi\ln\pars{2}}}
\end{align}