Chapter 7.8, Problem 68E

Improper Integrals that Are Both Type 1 and Type 2 The integral ${\displaystyle {\int }_a^{\infty }f}(x)dx$ is improper because the interval $[a,\infty )$ is infinite. If $f$ has an infinite discontinuity at $a$ , then the integral is improper for a second reason. In this case we evaluate the integral by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows:

${\displaystyle {\int }_a^{\infty }f}(x)dx={\displaystyle {\int }_a^{c}f}(x)dx+{\displaystyle {\int }_c^{\infty }f}(x)dxc>a$

Evaluate the given integral if it is convergent.

68. ${\displaystyle {\int }_2^{\infty }\frac{1}{x\sqrt{x^2-4}}}dx$