Improper Integral: ∫(sin(x)+2)/x^2 from 2 to ∞ - Converge or Diverge?

[Ping427]

Homework Statement

∫(sin(x)+2)/x^2 from 2 to infinity. Determine if this improper integral converge or diverge.2. The attempt at a solution
lim(x→infinity)=∫(sin(x)+2)/x^2 from 2 to t.

I know that if the integral ends up to be an infinite number, this will be converge otherwise, it will be diverge. However, I couldn't find a way to integrate this function. When I tried to graph it, I see that as x approaches to infinity, the function is getting closer to zero and not equal to. Is that means that it is diverge since even though the function is getting smaller and smaller, the area is still increasing.

 

[Mark44]

Homework Statement

∫(sin(x)+2)/x^2 from 2 to infinity. Determine if this improper integral converge or diverge.2. The attempt at a solution
lim(x→infinity)=∫(sin(x)+2)/x^2 from 2 to t.

I know that if the integral ends up to be an infinite number, this will be converge otherwise, it will be diverge. However, I couldn't find a way to integrate this function. When I tried to graph it, I see that as x approaches to infinity, the function is getting closer to zero and not equal to. Is that means that it is diverge since even though the function is getting smaller and smaller, the area is still increasing.

$-1 + 2 \le \sin(x) + 2 \le 1 + 2$
Does that help?

 

[Ping427]

$-1 + 2 \le \sin(x) + 2 \le 1 + 2$
Does that help?

Can I do (−1+2)/x^2 ≤ sin(x)/x^2+2 ≤(1+2)/x^2, then integrate 1/x^2 and 3/x^2 from 2 to ∞?
Then the answer will be 1/2 ≤ ∫(sin(x)+2)/x^2 ≤ 3/2, therefore, it will be converge?

 

[Mark44]

Can I do (−1+2)/x^2 ≤ sin(x)/x^2+2 ≤(1+2)/x^2, then integrate 1/x^2 and 3/x^2 from 2 to ∞?
Then the answer will be 1/2 ≤ ∫(sin(x)+2)/x^2 ≤ 3/2, therefore, it will be converge?

Yes, that's the idea.

 

[Ping427]

Yes, that's the idea.

Thank you for helping!