Answer
\[ = - {e^{ - t}}\,\,\,\left( {{t^2} + 2t + 2} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{t^2}\,{e^{ - t}}\,dt} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u\, = {t^2} \to du = 2tdt \hfill \\
dv = {e^{ - t}} \to v = - {e^{ - t}} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
= \frac{{{e^{ - t}}}}{{\,\left( { - 1} \right)}}{t^2} + \int_{}^{} {{e^{ - t}}\,\left( {2t} \right)} \,\,dt \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
= - {e^{ - t}}{t^2} + 2\int_{}^{} {{e^{ - t}}\,t} \,\,dt \hfill \\
\hfill \\
{\text{integrate}}\,\,{\text{by}}\,\,{\text{parts}}\,\,{\text{again}} \hfill \\
\hfill \\
= - {e^{ - t}}{t^2} + 2\,\,\left[ {\frac{{{e^{ - t}}}}{{\,\left( { - 1} \right)}} + t\int_{}^{} {{e^{ - t}}\,dt} } \right] \hfill \\
\hfill \\
= - {e^{ - t}}{t^2} - 2{e^{ - t}} \cdot t - 2{e^{ - t}} \hfill \\
\hfill \\
solution \hfill \\
\hfill \\
= - {e^{ - t}}\,\,\,\left( {{t^2} + 2t + 2} \right) + C \hfill \\
\end{gathered} \]
