## Calculus: Early Transcendentals (2nd Edition)

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