Answer
$$A = \frac{1}{e}$$
Work Step by Step
$$\eqalign{
& y = \left( {\ln x - 1} \right)/{x^2},{\text{ }}x \geqslant e \cr
& {\text{The area is given by }} \cr
& A = \int_e^{ + \infty } {\frac{{\ln x - 1}}{{{x^2}}}} dx \cr
& {\text{Using the definition of improper integrals}} \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \int_e^b {\left[ {\frac{{\ln x}}{{{x^2}}} - \frac{1}{{{x^2}}}} \right]} dx \cr
& {\text{Integrate }}\int {\frac{{\ln x}}{{{x^2}}}dx{\text{ by parts}}} \cr
& u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = \frac{1}{{{x^2}}}dx,{\text{ }}v = - \frac{1}{x} \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{1}{x}\ln x - \int {\left( { - \frac{1}{x}} \right)\left( {\frac{1}{x}} \right)} dx \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{1}{x}\ln x - \frac{1}{x} + C \cr
& {\text{Therefore,}} \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \int_e^b {\left[ {\frac{{\ln x}}{{{x^2}}} - \frac{1}{{{x^2}}}} \right]} dx \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \left[ { - \frac{1}{x}\ln x - \frac{1}{x} + \frac{1}{x}} \right]_e^b \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \left[ { - \frac{{\ln x}}{x}} \right]_e^b \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{{\ln e}}{e} - \frac{{\ln b}}{b}} \right] \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{e} - \frac{{\ln b}}{b}} \right] \cr
& {\text{Evaluate the limit}} \cr
& A = \frac{1}{e} - \frac{\infty }{\infty } \cr
& A = \frac{1}{e} - \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{b}} \right] \cr
& A = \frac{1}{e} \cr} $$
