<s>[INST] Show that $\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}$ converges. Show that 
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges.
I recognized that that since the integrand is even then
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} = 2\int_{0}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
Then, I started with the following statement and manipulated it to match the integrand.
$$
\begin{align*}
x^4 +1 &\gt x^4 \\ \sqrt{x^4+1} &\gt x^2 \\ \frac{1}{\sqrt{x^4 +1}} &\lt\frac{1}{x^2}
\end{align*}
$$
It is a well-known result that 
$$
\int_1^{\infty}\frac{dx}{x^2} 
$$
converges. So would it be correct to say that, by comparison, that 
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges as well?
 [/INST] In the regime $|x| \ge 1$, you can compare $(x^4+1)^{-1/2}$ against $x^{-2}$, but in the neighborhood of $0$, the latter is unbounded so you are best off noting that $$\int_{x=-\infty}^\infty (x^4+1)^{-1/2} \, dx = 2 \int_{x=0}^\infty (x^4+1)^{-1/2} \, dx < 2\left( \int_{x=0}^1 (x^4+1)^{-1/2} \, dx  + \int_{x=1}^\infty x^{-2} \, dx \right).$$  The first term is obviously finite since $0 < (x^4 + 1)^{-1/2} \le 1$ on $[0,1]$; and the second is also finite as you stated.
</s>