Answer
Convergent
Work Step by Step
$\displaystyle \int_0^{\infty}\frac{\arctan x}{2+e^x}\ dx$
Note the following:
$\displaystyle \frac{\arctan x}{2+e^x} < \frac{\arctan x}{e^x} \le \frac{\pi}{2e^x} < \frac{\pi}{2x^2}$
From the textbook, we know that $\displaystyle \int_0^{\infty}\frac{1}{x^p}\ dx$ converges when $\displaystyle p>1$.
Let's split the question into two integrals:
$\displaystyle \int_0^{\infty}\frac{\arctan x}{2+e^x}\ dx = \int_0^{1}\frac{\arctan x}{2+e^x}\ dx + \int_1^{\infty}\frac{\arctan x}{2+e^x}\ dx$
$\displaystyle \int_0^{1}\frac{\arctan x}{2+e^x}\ dx$ converges to some value, and since we do not need to know the value we focus on the convergence of $\displaystyle \int_1^{\infty}\frac{\arctan x}{2+e^x}\ dx$.
Above, we showed that $\displaystyle \frac{\pi}{2x^2}$ is larger than $\displaystyle \frac{\arctan x}{2+e^x}$ when $\displaystyle x \ge 1$, and $\displaystyle \frac{\pi}{2}\int_1^{\infty}\frac{1}{x^2}\ dx$ converges to a value. Therefore, by the comparison theorem, the lesser-valued $\displaystyle \int_1^{\infty}\frac{\arctan x}{2+e^x}\ dx$ also converges.
As a result, the integral in question is convergent.
