Answer
\[\frac{1}{4}\ln \left| t \right| + \frac{{{t^3}}}{6} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{1 + 2{t^3}}}{{4t}}dt} \hfill \\
rewrite\,\,the\,\,integrand\,\,as\,\,follows \hfill \\
\int_{}^{} {\,\left( {\frac{1}{{4t}} + \frac{{2{t^3}}}{{4t}}} \right)dt} \hfill \\
\int_{}^{} {\frac{1}{{4t}}dt + \int_{}^{} {\frac{{{t^2}}}{2}dt} } \hfill \\
Use\,\,indefinite\,\,integrals \hfill \\
\int_{}^{} {{t^{ - 1}}dt = \ln \left| t \right| + C\,\,,\,\,\int_{}^{} {{t^n}dt} = \frac{{{t^{n + 1}}}}{{n + 1}} + C} \hfill \\
\frac{1}{4}\ln \left| t \right| + \frac{1}{2}\,\left( {\frac{{{t^3}}}{3}} \right) + C \hfill \\
Simplifying \hfill \\
\frac{1}{4}\ln \left| t \right| + \frac{{{t^3}}}{6} + C \hfill \\
\end{gathered} \]
