Is
$\frac{1}{x^{\frac{3}{4}}}<\frac{\sqrt{x}+2+\cos{x}}{x+2},\forall x>x_0$
enough to prove that
$\int_{1}^{\infty}\frac{\sqrt{x}+2+\cos{x}}{x+2}dx$
diverges?
What are the standard techniques which can be used in these type of problems?
Is
$\frac{1}{x^{\frac{3}{4}}}<\frac{\sqrt{x}+2+\cos{x}}{x+2},\forall x>x_0$
enough to prove that
$\int_{1}^{\infty}\frac{\sqrt{x}+2+\cos{x}}{x+2}dx$
diverges?
What are the standard techniques which can be used in these type of problems?
Yes. The technique used here is: if $0 \le f(x) \le g(x)$ and $\int_a^b f$ diverges, then $\int_a^b g$ also diverges. See if you can prove it!
For this particular exercise it suffices to use the inequality $ \frac{{\sqrt x + 2 + \cos x}}{{x + 2}} \ge \frac{1}{{x + 2}} $ to conclude that the integral diverges, noting that $\int_1^\infty {\frac{1}{{x + 2}}dx} = \int_3^\infty {\frac{1}{x}dx} = \infty $.
Yes. The integral
$\int_{x_0}^{x_1}x^{-3/4}\mathrm dx=4\left(x_1^{1/4}-x_0^{1/4}\right)$
diverges for $x_1\to\infty$, and your integral is bounded from below by this integral (for $x_0\ge1$); hence your integral diverges.
By the way, you're missing the $\mathrm dx$ at the end of the integrals.