Answer
\[ = \frac{{\,{{\left( {{x^4} + 16} \right)}^7}}}{{28}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{x^3}\,{{\left( {{x^4} + 16} \right)}^6}} \,dx \hfill \\
\hfill \\
set\,\,the\,\,substitution \hfill \\
\hfill \\
u = {x^4} + 16\,\,\,\,\,\,then\,\,\,\,du = 4{x^3}dx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {{x^3}\,{{\left( {{x^4} + 16} \right)}^6}} \,dx = \frac{1}{4} \cdot \int_{}^{} {4{x^3}\,{{\left( {{x^4} + 16} \right)}^6}dx} \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
= \frac{1}{4} \cdot \int_{}^{} {{u^6}\,du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
= \frac{1}{4} \cdot \frac{{{u^7}}}{7} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,u = {x^4} + 16 \hfill \\
\hfill \\
= \frac{{\,{{\left( {{x^4} + 16} \right)}^7}}}{{28}} + C \hfill \\
\end{gathered} \]
