I tried to find out if this integral converges and I am not sure with my solution $$\DeclareMathOperator{\arctg}{\operatorname{arctg}}\int_{1}^{\infty} \frac{\arctg(\sqrt{1 + x^{2}})}{x+3}dx$$
Can I say, that $\frac{\arctg(\sqrt{1 + x^{2}})}{x+3}$ equivalent to $\frac{\pi/2}{x+3}$, that is equivalent to $\frac{\pi/2}{x}$ and $\int_{1}^{\infty} \frac{\pi/2}{x}dx$ is a divergent integral, and that is the reason, why the original integral also diverges?
Yes that's correct, in a more formal way we should refer to the limit comparison test, that is for $x \to \infty$
$$ \frac{\frac{\arctan(\sqrt{1 + x^{2}})}{x+3}}{\frac1{x+3}} \to \frac \pi 2$$
from which we can conclude for the divergence.
As an alternative, we can use that
$$\arctan t = \frac \pi 2 -\arctan \frac 1t \ge \frac \pi 2-\frac1t$$
therefore
$$\int_{1}^{\infty} \frac{\arctan\left(\sqrt{1 + x^{2}}\right)}{x+3}dx\ge \int_{1}^{\infty} \frac{\frac \pi 2}{x+3}dx- \int_{1}^{\infty} \frac{1}{\left(\sqrt{1 + x^{2}}\right)(x+3)}dx$$
where the last one converges.