Answer
\[ = \frac{8}{9}\]
Work Step by Step
\[\begin{gathered}
\int_0^2 {\frac{2}{{{s^3} + 3{s^2} + 3s + 1}}\,ds} \hfill \\
\hfill \\
factor\,ing\,\,the\,\,denominator \hfill \\
\hfill \\
= \int_0^2 {\frac{2}{{\,{{\left( {s + 1} \right)}^3}}}\,ds} \hfill \\
\hfill \\
set\,\,s + 1\, = \,t\,\,\,\, \to \,\,\,ds = dt \hfill \\
\hfill \\
= \int_1^3 {\frac{2}{{{t^3}}}} dt\,\, \hfill \\
\hfill \\
\,integrate\,and\,use\,\,the\,\,ftc \hfill \\
\hfill \\
= \frac{{ - 2}}{2}\,\,\left[ {\frac{1}{{{t^2}}}} \right]_1^3 \hfill \\
\hfill \\
= - \,\left( {\frac{1}{9} - 1} \right) \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
= \frac{8}{9} \hfill \\
\end{gathered} \]
