Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{\ln x}}{{{x^2}}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& \int_1^\infty {\frac{{\ln x}}{{{x^2}}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{\ln x}}{{{x^2}}}dx} \cr
& \cr
& {\text{Integrating }}\int {{x^{ - 2}}\ln x} dx{\text{ by parts}} \cr
& u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = {x^{ - 2}},{\text{ }}v = - {x^{ - 1}} \cr
& \int {{x^{ - 2}}\ln x} dx = - {x^{ - 1}}\ln x - \int { - {x^{ - 1}}} \left( {\frac{1}{x}} \right)dx \cr
& = - \frac{1}{x}\ln x + \int {\frac{1}{{{x^2}}}} dx \cr
& = - \frac{1}{x}\ln x - \frac{1}{x} + C \cr
& \cr
& \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{\ln x}}{{{x^2}}}dx} = \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{x}\ln x - \frac{1}{x}} \right]_1^b \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{x}\ln x + \frac{1}{x}} \right]_1^b \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left[ {\left( {\frac{1}{b}\ln b + \frac{1}{b}} \right) - \left( {\frac{1}{1}\ln 1 + \frac{1}{1}} \right)} \right] \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left[ {\left( {\frac{1}{b}\ln b + \frac{1}{b}} \right) - 1} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& = - \left[ {\left( {\frac{1}{\infty }\ln \infty + \frac{1}{\infty }} \right) - 1} \right] \cr
& = - \left[ {\left( {0 + 0} \right) - 1} \right] \cr
& = 1 \cr} $$
