Answer
\[ = \frac{1}{9}\ln \left| {\frac{y}{{2y + 9}}} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dy}}{{y\,\left( {2y + 9} \right)}}} \hfill \\
\hfill \\
use\,\, \hfill \\
\hfill \\
\int_{}^{} {\frac{{du}}{{u\,\left( {bu + a} \right)}}} = \frac{1}{a}\ln \left| {\frac{u}{{bu + a}}} \right| + C \hfill \\
\hfill \\
b = 2,{\text{ }}a = 3 \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \frac{1}{9}\ln \left| {\frac{y}{{2y + 9}}} \right| + C \hfill \\
\end{gathered} \]
