Answer
\[ = \frac{{\,{{\left( {{x^6} - 3{x^2}} \right)}^5}}}{{30}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,{{\left( {{x^6} - 3{x^2}} \right)}^4}\,\left( {{x^5} - x} \right)dx} \hfill \\
\hfill \\
set\,\,the\,\,substitution \hfill \\
\hfill \\
u = {x^6} - 3{x^2}\,\,\,\,\,\,then\,\,\,\,\,\,du = 6{x^5} - 6xdx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\,{{\left( {{x^6} - 3{x^2}} \right)}^4}\,\left( {{x^5} - x} \right)dx} \, = \frac{1}{6} \cdot \int_{}^{} {\,{{\left( {{x^6} - 3{x^2}} \right)}^4}\,\left( {6{x^5} - 6x} \right)dx} \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
\frac{1}{6}\int_{}^{} {{u^4}du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
\frac{1}{6}.\frac{{{u^5}}}{5} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,\,u = {x^6} - 3{x^2} \hfill \\
\hfill \\
= \frac{{\,{{\left( {{x^6} - 3{x^2}} \right)}^5}}}{{30}} + C \hfill \\
\end{gathered} \]
