Answer
\[\begin{gathered}
\frac{3}{4}\,\left( {2u - 7\ln \left| {2u + 7} \right|} \right)\, + C \hfill \\
\hfill \\
\end{gathered} \]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{3u}}{{2u + 7}}} \,du \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
= 3\int_{}^{} {\frac{u}{{2u + 7}}} \,du \hfill \\
\hfill \\
use\,\,the\,\,formula\,\, \hfill \\
\hfill \\
\int_{}^{} {\frac{u}{{bu + a}}} = \frac{1}{{{b^2}}}\,\left( {bu - a\ln \left| {a + bu} \right|} \right) + C \hfill \\
\hfill \\
with\,\,a = 7\,\,b = 2 \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
3\int_{}^{} {\frac{u}{{2u + 7}}} \,du = \frac{3}{4}\,\left( {2u - 7\ln \left| {2u + 7} \right|} \right)\, + C \hfill \\
\hfill \\
\end{gathered} \]
