Answer
$1$
Work Step by Step
Table of integrals : 101. $\displaystyle \int u^{n}\ln udu=\frac{u^{n+1}}{(n+1)^{2}}[(n+1)\ln u-1]+C$
$\displaystyle \int x^{-2}\ln xdx=\frac{x^{-1}}{(-1)^{2}}[(-1)\ln x-1]+C =\displaystyle \frac{-\ln x-1}{x}+C$
This is a type I improper integral.
$\displaystyle \int_{1}^{\infty}\frac{\ln x}{x^{2}}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{\ln x}{x^{2}}dx=\lim_{t\rightarrow\infty}\left[\frac{-\ln x-1}{x}\right]_{1}^{t}$
$=\displaystyle \lim_{t\rightarrow\infty}\frac{-\ln t-1}{t}-(-1)$
...Apply L'Hospital's rule, $\displaystyle \frac{\infty}{\infty}$
$=\displaystyle \lim_{t\rightarrow\infty}\frac{-\frac{1}{t}}{1}+1$
$=0+1$
$=1$