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.
$$
\begin{align*}
x^4 +1 &amp;\gt x^4 \\ \sqrt{x^4+1} &amp;\gt x^2 \\ \frac{1}{\sqrt{x^4 +1}} &amp;\lt\frac{1}{x^2}
\end{align*}
$$

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?
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 $$\int_{x=-\infty}^\infty (x^4+1)^{-1/2} \, dx = 2 \int_{x=0}^\infty (x^4+1)^{-1/2} \, dx &lt; 2\left( \int_{x=0}^1 (x^4+1)^{-1/2} \, dx  + \int_{x=1}^\infty x^{-2} \, dx \right).$$  The first term is obviously finite since $0 &lt; (x^4 + 1)^{-1/2} \le 1$ on $[0,1]$; and the second is also finite as you stated.
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$\ds{\color{#f00}{\int_{-\infty}^{\infty}{\dd x  \over \root{1 + x^{4}}}} =
     2\int_{0}^{\infty}{\dd x  \over \root{1 + x^{4}}}}$. Let
$\ds{t \equiv {1 \over 1 + x^{4}}}$ such that
$\ds{x = \pars{{1 \over t} - 1}^{1/4}}$.

Then,

\begin{align}
\ds{\color{#f00}{\int_{-\infty}^{\infty}{\dd x  \over \root{1 + x^{4}}}}} &amp; =
2\int_{1}^{0}t^{1/2}\,{1 \over 4}\,\pars{{1 \over t} - 1}^{-3/4}
\pars{-\,{1 \over t^{2}}}\,\dd t
\\[3mm] &amp; = \half\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-3/4}\,\dd t =
\half\,{\Gamma\pars{1/4}\Gamma\pars{1/4} \over \Gamma\pars{1/4 + 1/4}}
\\[3mm] &amp; =
\color{#f00}{{1 \over 2\root{\pi}}\,\Gamma^{2}\pars{{1 \over 4}}} \approx
3.7081
\end{align}

since $\Gamma\pars{1/2} = \root{\pi}$.