Converging or diverging integral

We will use the comparison theorem to prove this. Let $f(x) =\frac{\arctan{(x)}}{e^{x}+1}$ We know the maximum value $\arctan{(x)}$ can take is $\frac{\pi}{2}$ , therefore we can set an upper bound of $f(x)$ as $\frac{\pi}{2e^x + 2}$ . If we can prove that this is less than a function $g(x)$ for all $x>0$ then we have consequently shown that $f(x) < g(x)$ for all $x>0$ . If this is the case, and if $g(x)$ converges, then $f(x)$ must also converge. We now let $g(x) = \frac{\pi}{e^{x}}$ , since we know $\int_{0}^{\infty}\frac{\pi}{e^{x}} = \pi$ . Now with a little rearranging, it is clear that $\frac{\pi}{2e^x + 2} < \frac{\pi}{e^x}$ and therefore the integral must converge.