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: 
Let $I$ be the integral given by 

$$\begin{align}
I&=\int\_0^\infty\frac{\log(1+x^2)}{1+x^2}\,dx\\\\
\end{align}$$

Let $x=\tan(\theta)$. Then, 

$$\begin{align}
I&=-2\int\_0^{\pi/2} \log(\cos(\theta))\,d\theta\\\\
&=-2\int\_0^{\pi/2}\log(\sin(\theta))\\\\
&=\pi\log(2)
\end{align}$$

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To see why $\int\_0^{\pi/2}\log(\sin(\theta))\,d\theta=-\pi\log(2)$, simply note that 

$$\begin{align}
\int\_0^{\pi/2}\log(\sin(\theta))\,d\theta&=\frac12\int\_0^{\pi/2}(\log(\sin(\theta))+\log(\cos(\theta)))\,d\theta\\\\
&=\frac12\int\_0^{\pi/2}\log\left(\frac12\sin(2\theta)\right)\,d\theta\\\\
&=-\frac{\pi}{4}\log(2)+\frac12\int\_0^{\pi/2}\log\left(\sin(2\theta)\right)\,d\theta\\\\
&=-\frac{\pi}{4}\log(2)+\frac14\int\_0^{\pi}\log\left(\sin(\theta)\right)\,d\theta\\\\
&=-\frac{\pi}{4}\log(2)+\frac12\int\_0^{\pi/2}\log\left(\sin(\theta)\right)\,d\theta\\\\
\frac12\int\_0^{\pi/2}\log\left(\sin(\theta)\right)\,d\theta&=-\frac{\pi}{4}\log(2)\\\\
\int\_0^{\pi/2}\log\left(\sin(\theta)\right)\,d\theta&=-\frac{\pi}{2}\log(2)
\end{align}$$
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.