Answer
\[\frac{1}{3}\ln \left| {{t^3} + 6t + 3} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{t^2} + 2}}{{{t^3} + 6t + 3}}dt} \hfill \\
Let\,\,u = {t^3} + 6t + 3\,\,,\,So\,\,that \hfill \\
du = \,\left( {3{t^2} + 6} \right)dt \hfill \\
du = 3\,\left( {{t^2} + 2} \right)dt \hfill \\
\int_{}^{} {\frac{{{t^2} + 2}}{{{t^3} + 6t + 3}}dt\,\, = \frac{1}{3}\int_{}^{} {\frac{{3\,\left( {{t^2} + 2} \right)dt}}{{{t^3} + 6t + 3}}} } \hfill \\
\frac{1}{3}\int_{}^{} {\frac{{du}}{u}} \hfill \\
Integrating \hfill \\
\frac{1}{3}\ln \left| u \right| + C \hfill \\
Substituting\,\,u = {t^3} + 6t + 3\,\,gives \hfill \\
\frac{1}{3}\ln \left| {{t^3} + 6t + 3} \right| + C \hfill \\
\end{gathered} \]
