Answer
Converges
Work Step by Step
Use the limit comparison test with $\dfrac{1}{x^2+1}$:
$\lim\limits_{x \to \infty} \dfrac{1/\sqrt {x^4+1}}{1/(x^2+1)}=\sqrt {\lim\limits_{x \to \infty} \dfrac{x^4+2x^2+1}{x^4+1}}=1$
Now, $\int_{-\infty}^{\infty} \dfrac{dx}{x^2+1}=\int_{-\infty}^{0} \dfrac{dx}{x^2+1} +\int_{0}^{\infty} \dfrac{dx}{x^2+1} \\=2 \int_{0}^{\infty} \dfrac{dx}{x^2+1} \\=2 \lim\limits_{a \to \infty}\int_{0}^{a} \dfrac{dx}{x^2+1} \\= 2 \lim\limits_{a \to \infty}[\tan^{-1} x]_{0}^a=2 (\dfrac{\pi}{2}-0) \\=\pi$
Thus, the integral converges by limit comparison test.