Answer
=1
Work Step by Step
\[\begin{gathered}
\int_1^\infty {\frac{{\ln x}}{{{x^2}}}} \,dx \hfill \\
\hfill \\
use\,\,the\,\,Formula\,\,for\,\,in\,tegration\,\,by\,\,parts \hfill \\
\hfill \\
\int_{}^{} {udv = uv - \int_{}^{} {vdu} } \hfill \\
\hfill \\
set \hfill \\
du = \ln x\,\,\,\,\,\,\,then\,\,\,\,du = \frac{1}{x}dx \hfill \\
dv = \frac{1}{{{x^2}}}dx\,\,then\,\,\,\,v = - \frac{1}{x} \hfill \\
\hfill \\
integrating \hfill \\
\hfill \\
\int_{}^{} {\frac{{\ln x}}{{{x^2}}}} \,dx = \,\,\left[ { - \frac{1}{x}\ln x - \frac{1}{x}} \right] \hfill \\
\hfill \\
\int_1^\infty {\frac{{\ln x}}{{{x^2}}}} \,dx\,\, = \mathop {\lim }\limits_{a \to 0} \int_1^a {\frac{{\ln x}}{{{x^2}}}\,dx} \hfill \\
\hfill \\
= \mathop {\lim }\limits_{a \to 0} \,\,\left[ { - \frac{1}{x}\ln x - \frac{1}{x}} \right]_1^a \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \mathop {\lim }\limits_{a \to 0} \,\,\left[ { - \frac{1}{a}\ln a - \frac{1}{a} + 1} \right] \hfill \\
\hfill \\
evaluate\,\,the\,\,\,limit \hfill \\
\hfill \\
= 1 \hfill \\
\hfill \\
\end{gathered} \]
