Answer
\[ = \infty \]$$\boxed{\textbf{(Diverges.)}}$$
Work Step by Step
\[\begin{gathered}
\int_2^\infty {\frac{{dy}}{{y\ln y}}} \hfill \\
\hfill \\
set \hfill \\
u = \ln y\,\,\,\,\,then\,\,\,du = \,{\left( {\ln y} \right)^\prime }dy \hfill \\
and\,\,du = \frac{1}{y}dy \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_2^\infty {\frac{{dy}}{{y\ln y}}} = \int_2^\infty {\frac{{dy}}{{\ln y}}} \,\left( {\frac{1}{y}dy} \right) \hfill \\
\hfill \\
= \int_{\ln 2}^{\ln \,\infty } {\frac{1}{u}\,\,du} = \int_{\ln 2}^\infty {\frac{1}{u}du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \mathop {\lim }\limits_{b \to \infty } \int_{\ln 2}^b {\frac{1}{u}du} = \mathop {\lim }\limits_{b \to \infty } \,\,\left[ {\ln u} \right]_{\ln 2}^b \hfill \\
\hfill \\
evaluate\,\,the\,\,{\text{limits}}\,\,and\,\,simplify \hfill \\
\hfill \\
= \mathop {\lim }\limits_{b \to \infty } \,\,\left[ {\ln b - \ln \,\left( {\ln 2} \right)} \right] \hfill \\
\hfill \\
= \infty - \ln \,\left( {\ln 2} \right) = \infty \hfill \\
diverges \hfill \\
\hfill \\
\end{gathered} \]
