Check if $\int _0^{\infty }\:\left(\frac{\arctan\left(x\right)}{\:2+e^{3x}}\right)dx$ converges/diverges. $$\text{Does }\int _0^{\infty }\:\left(\frac{\arctan\left(x\right)}{\:2+e^{3x}}\right)dx \text { converge/diverge?}$$
I have to use the convergence/divergence theorem to check if that integral converges or diverges.
I notice that:
$$\frac{\arctan(x)}{2+e^{3x}} \leq \frac{\arctan(x)}{e^{3x}} \leq \frac{\arctan(x)}{e^{x}}$$
I also notice that $$-\pi/2 \leq \arctan(x) \leq \pi/2$$.
So I can say that 
$$\frac{\arctan(x)}{2+e^{3x}} \leq \frac{\arctan(x)}{e^{3x}} \leq \frac{\arctan(x)}{e^{x}} \ \leq \frac{\pi/2}{e^{x}}$$
So here, would I just prove that $$\int^{\infty}_{0}\frac{\pi/2}{e^x}dx$$ converges?
Could I also take it one more step and say that 
$$\frac{\pi/2}{e^x} \leq {\pi/2}$$
since we want to maximize the numerator while making the denominator smaller, and we know that $\forall x \in [0,\infty), 1\leq e^x \leq \infty$?
So in the end, can I just prove that $\int^{\infty}_{0} \pi/2 dx$ converges?
 A: There are a couple of issues in the OP that need to be addressed.  First, we should bound the integrand as 
$$\left|\frac{\arctan(x)}{2+e^{3x}}\right|\le \frac{\pi/2}{2+e^{3x}}$$
Note the absolute value sign prevents the possibility of the integrand from running to far negative.  In this problem, that isn't an issue inasmuch as the integrand is non-negative on the domain of integration.


The first question from the OP is:
"So I can say that 
$$\frac{\arctan(x)}{2+e^{3x}} \leq \frac{\arctan(x)}{e^{3x}} \leq \frac{\arctan(x)}{e^{x}} \ \leq \frac{\pi/2}{e^{x}}"$$
">"So here, would I just prove that $$\int^{\infty}_{0}\frac{\pi/2}{e^x}dx$$ converges?"

Yes, the inequalities are correct and one can proceed to show 
$$\begin{align}
\left|\int_0^\infty \frac{\arctan(x)}{2+e^{3x}}\,dx\right|&\le \frac{\pi}{2}\int_0^\infty e^{-x}\,dx\\\\
&=1\\\\
&<\infty
\end{align}$$
And we are done!


The second question from the OP is:
"Could I also take it one more step and say that 
$$\frac{\pi/2}{e^x} \leq {\pi/2}$$
  since we want to maximize the numerator while making the denominator smaller, and we know that $\forall x \in [0,\infty), 1\leq e^x \leq \infty$?"
"So in the end, can I just prove that $\int^{\infty}_{0} \pi/2 dx$ converges?"

No.  While it is true that $\frac{\pi/2}{e^x}\le \frac{\pi}{2}$ for $x\in[0,\infty)$, this upper bound provides no insight into the convergence or lack thereof of the integral of interest.  That is to say, it shows
$$\left|\int_0^L\frac{\arctan(x)}{2+e^{3x}}\,dx\right| \le \int_0^L \frac{\pi}{2}\,dx\to  \infty\,\,\text{as}\,\,L\to \infty \tag 1$$
So, from $(1)$ we could not deduce that the integral of interest converges.
A: You are very close
when you write
"So here, would I just prove that $$\int^{\infty}_{0}\frac{\pi/2}{e^x}dx$$ converges?"
And the answer is yes.
Since
$(e^{-x})' = -e^{-x}$,
$\int e^{-x} dx
= -e^{-x}$,
so
$\int_0^{\infty} e^{-x} dx
= (-e^{-x})\big|_0^{\infty}
=1$,
so it converges.
