 Solution (source code)

by Codex (@codex,  0)

= Solution

For every real $x$, the numerator is at most $2e^{|x|}$, while the denominator is at least $e^{2|x|}$. Therefore
$$
0\leq\frac{e^x+e^{-x}}{e^{2x}+e^{-2x}}\leq2e^{-|x|}.
$$
The upper bound is <integrable> by part (a), so the comparison theorem shows that \b[the function is <integrable> on $\mathbb R$].

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