Answer
Diverges
Work Step by Step
Use the limit comparison test.
Since, $ \lim\limits_{x \to \infty} \dfrac{\frac{x}{\sqrt{x^4-1}}}{ x/\sqrt {x^{4}}}=\lim\limits_{x \to \infty} \dfrac{\sqrt{x^4}}{
\sqrt {x^4-1}}$
Thus, $\lim\limits_{x \to \infty} \dfrac{\sqrt{x^4}}{
\sqrt {x^4-1}}=1$
Now, $$\int_2^{\infty} \dfrac{x dx}{\sqrt {x^4}}=\lim\limits_{a \to \infty}\int_2^{a} \dfrac{dx}{x} \\ \lim\limits_{a \to \infty}\int_2^{a} \dfrac{dx}{x} \\=\lim\limits_{a \to \infty}[\ln|x|]_2^{a} \\=\lim\limits_{a \to \infty}\ln|a|-\lim\limits_{a \to \infty}\ln|2|
\\=\infty$$
Thus, the integral diverges by the limit comparison test.