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My attempt: We know $f(x)=\frac{\cos^2x}{\sqrt{x^4+1}}$ (or rather, I am just assuming) is locally integrable and positive, and $$0\leq f(x)\leq\frac{1}{\sqrt{x^4+1}}\leq\frac{1}{x^2},$$ and we know that $\frac{1}{x^2}$ is improperly integrable on $(0,\infty)$, so by the comparison theorem for improper integrals, $f(x)$ is improperly integrable.

I am fairly sure that I am supposed to use the comparison theorem for improper integrals, and that this is roughly the argument that I was supposed to make, but I am not completely sure if I correctly made my argument.

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  • $\begingroup$ There are problems on the $0$ edge of the region of integration if you try just comparing to $1/x^2$ (See P.Zhou's answer) $\endgroup$ – Peter Woolfitt Dec 15 '15 at 5:40
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For $0<x<1$, bound the integrand by $1$; for $x>1$, bound the integrand by $1/x^2$, which is integrable.

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