Comparison Theorem for Integral Calculus

Question

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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.

Answer 1

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).