For which values of the parameter $\beta$ is the following integral convergent For which values of the parameter $\beta$ is the following integral convergent
$$\int_0^\infty \frac{3arctan(x)}{(1+x^{\beta-1})(2+cos(x))}\,dx$$
I tried using comparison test with $g(x) = \frac{1}{x^{\beta-1}}$, however I realized that for example for $\beta-1 = \frac{1}{2}$ it won't work and my idea got 'destroyed'
What should be the proper approach to this problem?

Note that the integrand has no singularities on the positive real axis, so your only concern is the behaviour at $+\infty$. As the integrand is non-negative, the integral converges if and only if
$$ f(\beta) \equiv \int \limits_1^\infty \frac{3 \arctan (x)}{(1+x^{\beta-1})(2+\cos(x))} \, \mathrm{d} x < \infty $$
holds. For $x \in (1,\infty)$ we have $ \arctan(x) \in (\frac{\pi}{4},\frac{\pi}{2})$ and $2+\cos(x) \in [1,3]$, so we get the estimate
$$ \frac{\pi}{4} \int \limits_1^\infty \frac{1}{1+x^{\beta-1}} \, \mathrm{d} x\leq f(\beta) \leq \frac{3 \pi}{2} \int \limits_1^\infty \frac{1}{1+x^{\beta-1}} \, \mathrm{d} x \leq \frac{3 \pi}{2} \int \limits_1^\infty x^{1-\beta} \, \mathrm{d} x
$$
for any $\beta \in \mathbb{R}$.
Now if $\beta \leq 1$, we have $x^{\beta-1} \leq 1$ for $x\geq1$, so the estimate from below yields
$$ f(\beta) \geq \frac{\pi}{8} \int \limits_1^\infty 1 \, \mathrm{d} x = \infty$$
and the integral diverges. If $\beta > 1$, we can use $x^{\beta -1} \geq 1$ for $x \geq 1$ to find
$$ \frac{\pi}{8} \int \limits_1^\infty x^{1-\beta} \, \mathrm{d} x\leq f(\beta) \leq \frac{3 \pi}{2} \int \limits_1^\infty x^{1-\beta} \, \mathrm{d} x \, .
$$
