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?

$$
\begin{align}
\int_{-\infty}^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}
&=2\int_0^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag1\\
&=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_1^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag2\\
&=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag3\\
&=4\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag4
\end{align}
$$
Explanation:
$(1)$: symmetry
$(2)$: split the integral into two parts
$(3)$: substitute $x\mapsto\frac1x$
$(4)$: distribute
Because the integrand is between $\frac1{\sqrt2}$ and $1$, the integral converges and is between $2\sqrt2$ and $4$.
