Answer
\[ = \frac{3}{5}\,{\left( {x + 4} \right)^{\frac{5}{3}}} - 6\,{\left( {x + 4} \right)^{\frac{2}{3}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{x}{{\sqrt[3]{{x + 4}}}}\,dx} \hfill \\
\hfill \\
Set\,\,\,u = x + 4\,\,,\,\,so \hfill \\
\hfill \\
x = u - 4\,\,\,\,\,{\text{which implies that}}\,\,\,\,dx = du\,\, \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \int_{}^{} {\frac{{u - 4}}{{{u^{\frac{1}{3}}}}}\,du} \hfill \\
\hfill \\
distribute \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {{u^{\frac{2}{3}}} - 4{u^{ - \frac{1}{3}}}} \right)} \,du \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{3}{5}{u^{\frac{5}{3}}} - 6{u^{\frac{2}{3}}} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,\,u = x + 4 \hfill \\
\hfill \\
= \frac{3}{5}\,{\left( {x + 4} \right)^{\frac{5}{3}}} - 6\,{\left( {x + 4} \right)^{\frac{2}{3}}} + C \hfill \\
\hfill \\
\end{gathered} \]