Answer
\[ = - \frac{s}{2}{e^{ - 2s}} - \frac{1}{4}{e^{ - 2s}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {s{e^{ - 2s}}ds} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,du = ds \hfill \\
dv = {e^{ - 2s}}\,\,ds\,\,\,\, \to \,\,\,\,\,v\,\, = - \frac{1}{2}{e^{ - 2s}} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
= - \frac{s}{2}{e^{ - 2s}} + \int_{}^{} {\frac{1}{2}{e^{ - 2s}}ds} \hfill \\
then \hfill \\
= - \frac{s}{2}{e^{ - 2s}} - \frac{1}{4}\int_{}^{} {{e^{ - 2s}}\left( { - 2} \right)ds} \hfill \\
\hfill \\
{\text{integrating}} \hfill \\
\hfill \\
= - \frac{s}{2}{e^{ - 2s}} - \frac{1}{4}{e^{ - 2s}} + C \hfill \\
\end{gathered} \]
