Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_1^{ + \infty } {\frac{1}{{{x^{1.5}}}}dx} = \int_1^{ + \infty } {{x^{ - 1.5}}dx} \cr
& {\text{Using the definition of improper integrals see page 1053}} \cr
& \underbrace {\int_a^{ + \infty } {f\left( x \right)dx = \mathop {\lim }\limits_{M \to + \infty } } \int_a^M {f\left( x \right)} dx}_ \Downarrow \cr
& \int_1^{ + \infty } {{x^{ - 1.5}}dx} = \mathop {\lim }\limits_{M \to + \infty } \int_1^M {{x^{ - 1.5}}dx} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{M \to + \infty } \left[ {\frac{{{x^{ - 0.5}}}}{{ - 0.5}}} \right]_1^M \cr
& = - 2\mathop {\lim }\limits_{M \to + \infty } \left[ {\frac{1}{{{x^{0.5}}}}} \right]_1^M \cr
& = - 2\mathop {\lim }\limits_{M \to + \infty } \left[ {\frac{1}{{\sqrt x }}} \right]_1^M \cr
& = - 2\mathop {\lim }\limits_{M \to + \infty } \left[ {\frac{1}{{\sqrt M }} - 1} \right] \cr
& {\text{Evaluate the limit when }}M \to + \infty \cr
& = - 2\left[ {\frac{1}{{\sqrt { + \infty } }} - 1} \right] \cr
& = - 2\left[ {0 - 1} \right] \cr
& = 2 \cr} $$
