Answer
\[ = \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + t - 3\ln \,\left| {t + 1} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{t^3} - 2}}{{t + 1}}dt} \hfill \\
\hfill \\
{\text{using}}\,\,{\text{the}}\,\,{\text{long}}\,\,{\text{division}} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {{t^2} - t + 1 - \frac{3}{{t + 1}}} \right)dt} \hfill \\
\hfill \\
Therefore, \hfill \\
\hfill \\
= \int_{}^{} {\,{t^2}dt} - \int_{}^{} {\,tdt} + \int_{}^{} {\,dt} - \int_{}^{} {\frac{3}{{t + 1}}\,dt} \hfill \\
\hfill \\
integrate\, \hfill \\
\hfill \\
= \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + t - 3\ln \,\left| {t + 1} \right| + C \hfill \\
\end{gathered} \]
