Answer
The integral converges.
Work Step by Step
We use the limit comparison test as follows:
$\lim\limits_{x \to \infty} \dfrac{1/\sqrt {x^4+1}}{1/(x^2+1)}=\sqrt {\lim\limits_{x \to \infty} \dfrac{x^4+2x^2+1}{x^4+1}}=1$
Since we have a non-zero finite limit, then the functions will either both converge or diverge. We need to test $g(x)$ for convergence.
So, $\int_{-\infty}^{\infty} \dfrac{dx}{x^2+1}=\int_{-\infty}^{0} \dfrac{dx}{x^2+1} +\int_{0}^{\infty} \dfrac{dx}{x^2+1}$
$2 \int_{0}^{\infty} \dfrac{dx}{x^2+1}=2 \lim\limits_{a \to \infty}\int_{0}^{a} \dfrac{dx}{x^2+1}= 2 \lim\limits_{a \to \infty}[\tan^{-1} x]_{0}^a=2 (\dfrac{\pi}{2}-0)=\pi$
As the integral of the function $g(x)$ converges, then the integral converges as well.
