Answer
\[\frac{{{e^{2u}}}}{2} + 2{u^2} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,\left( {{e^{2u}} + 4u} \right)} du \hfill \\
By\,\,the\,\,sum\,\,and\,\,difference\,\,rule \hfill \\
\int_{}^{} {{e^{2u}}du + \int_{}^{} {4udu} } \hfill \\
Use\,\,indefinite\,\,integrals \hfill \\
\int_{}^{} {{e^{kx}}dx = \frac{{{e^{kx}}}}{k} + C\,\,,\,\,\int_{}^{} {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} } \hfill \\
Then \hfill \\
\frac{{{e^{2u}}}}{2} + 4\,\left( {\frac{{{u^2}}}{2}} \right) + C \hfill \\
Simplifying \hfill \\
\frac{{{e^{2u}}}}{2} + 2{u^2} + C \hfill \\
\end{gathered} \]
