Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_1^{ + \infty } {\frac{{\ln x}}{{{x^2}}}} dx \cr
& {\text{Integrate }}\int {\frac{{\ln x}}{{{x^2}}}dx{\text{ by parts}}} \cr
& u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = \frac{1}{{{x^2}}}dx,{\text{ }}v = - \frac{1}{x} \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{1}{x}\ln x - \int {\left( { - \frac{1}{x}} \right)\left( {\frac{1}{x}} \right)} dx \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{1}{x}\ln x - \frac{1}{x} + C \cr
& {\text{Therefore,}} \cr
& \int_1^{ + \infty } {\frac{{\ln x}}{{{x^2}}}} dx = \mathop {\lim }\limits_{b \to + \infty } \left[ { - \frac{1}{x}\ln x - \frac{1}{x}} \right]_1^b \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to + \infty } \left[ {\left( { - \frac{1}{b}\ln b - \frac{1}{b}} \right) - \left( { - \frac{1}{1}\ln 1 - \frac{1}{1}} \right)} \right] \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to + \infty } \left[ {\left( { - \frac{1}{b}\ln b - \frac{1}{b}} \right) + 1} \right] \cr
& {\text{ }} = - \mathop {\lim }\limits_{b \to + \infty } \frac{1}{b}\ln b - \mathop {\lim }\limits_{b \to + \infty } \frac{1}{b} + \mathop {\lim }\limits_{b \to + \infty } 1 \cr
& {\text{ }} = - \mathop {\lim }\limits_{b \to + \infty } \frac{{\ln b}}{b} - \mathop {\lim }\limits_{b \to + \infty } \frac{1}{b} + \mathop {\lim }\limits_{b \to + \infty } 1 \cr
& {\text{By the LHopital's rule}} \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to + \infty } \frac{{\ln b}}{b} = \mathop {\lim }\limits_{b \to + \infty } \frac{{1/b}}{1} = 0 \cr
& {\text{Then}} \cr
& {\text{ }} = - \mathop {\lim }\limits_{b \to + \infty } \frac{{\ln b}}{b} - \mathop {\lim }\limits_{b \to + \infty } \frac{1}{b} + \mathop {\lim }\limits_{b \to + \infty } 1 \cr
& {\text{ }} = 0 - 0 + 1 \cr
& {\text{ }} = 1 \cr
& \int_1^{ + \infty } {\frac{{\ln x}}{{{x^2}}}} dx = 1 \cr} $$
