Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.8 Exercises - Page 575: 46

${\text{Diverges}}$

$$\eqalign{ & \int_4^\infty {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}} dx \cr & {\text{By the Definition of Improper Integrals with Infinite }} \cr & {\text{Integration Limits}} \cr & \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr & \int_4^\infty {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_4^b {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}dx} \cr & {\text{Integrate by tables using:}} \cr & \int {\frac{{\sqrt {{u^2} - {a^2}} }}{{{u^2}}}du} = - \frac{{\sqrt {{u^2} - {a^2}} }}{u} + \ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C \cr & {\text{Let }}u = x{\text{ and }}a = 4 \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{b \to \infty } \int_4^b {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}dx} = \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{{\sqrt {{x^2} - 16} }}{x} + \ln \left| {x + \sqrt {{x^2} - 16} } \right|} \right]_4^b \cr & = \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{{\sqrt {{b^2} - 16} }}{b} + \ln \left| {b + \sqrt {{b^2} - 16} } \right|} \right] \cr & - \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{{\sqrt {{4^2} - 16} }}{4} + \ln \left| {4 + \sqrt {{4^2} - 16} } \right|} \right] \cr & = \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{{\sqrt {{b^2} - 16} }}{b} + \ln \left| {b + \sqrt {{b^2} - 16} } \right|} \right] - \mathop {\lim }\limits_{b \to \infty } \left[ { - 0 + \ln \left| 4 \right|} \right] \cr & {\text{Evaluate the limit when }}b \to \infty \cr & = \mathop {\lim }\limits_{b \to \infty } \left[ { - 1 + \ln \left| \infty \right|} \right] - \ln \left| 4 \right| \cr & = \infty \cr & {\text{So}},{\text{ the improper integral diverges}}. \cr} $$