Comparison Theorem for Integral Calculus

I have narrowed it down to C, E, and F, since we know that $1/x^{1/5}$ is always greater than the original function for all $x\geq 1$. However, the second set of conditions is more difficult to understand.

-
I'm still trying to understand what it means, sorry...How can I change it? –  EngGenie Jan 16 '13 at 3:05
go to your answered questions and choose the answer that you liked better. You can upvote many answers in each question you ask, but you can "accept" only one. –  DonAntonio Jan 16 '13 at 3:06
Looks nicer now, right? –  EngGenie Jan 16 '13 at 3:07
Indeed it does. Way to go, most people appreciates that kind of things here. –  DonAntonio Jan 16 '13 at 3:11

If $x \ge 1$, then $(x + e^{2x})^{\frac{1}{5}} \ge (e^{2x})^{\frac{1}{5}} = e^{\frac{2x}{5}}$. So $(x+e^{2x})^{-\frac{1}{5}}\le e^{-\frac{2x}{5}}$.
$\displaystyle \int_1^{\infty} e^{-\frac{2x}{5}} dx = \frac{5}{2}e^{-\frac{2}{5}} < \infty$. So it's f).