## University Calculus: Early Transcendentals (3rd Edition)

Apply the limit comparison test. Thus, we have $\lim\limits_{x \to \infty} \dfrac{x/\sqrt{x^4-1}}{ x/\sqrt {x^{4}}}=\lim\limits_{x \to \infty} \dfrac{\sqrt{x^4}}{ \sqrt {x^4-1}}$ This suggests that $\lim\limits_{x \to \infty} \dfrac{\sqrt{x^4}}{ \sqrt {x^4-1}}=1$ Since we got a finite positive value, we know that both series converge or both diverge. This implies that $\int_2^{\infty} \dfrac{x dx}{\sqrt {x^4}}=\lim\limits_{p \to \infty}\int_2^{p} \dfrac{dx}{x}$ and $\lim\limits_{p \to \infty}\int_2^{p} \dfrac{dx}{x}=\lim\limits_{p \to \infty}[\ln|x|]_2^{p} =\infty$ Hence, the given integral diverges by the limit comparison test.