Prove

Let $x^{3}-1=t$

$\therefore 3 x^{2} d x=d t$

$\Rightarrow \int\left(x^{3}-1\right)^{\frac{1}{3}} x^{5} d x=\int\left(x^{3}-1\right)^{\frac{1}{3}} x^{3} \cdot x^{2} d x$

$=\int t^{\frac{1}{3}}(t+1) \frac{d t}{3}$

$=\frac{1}{3} \int\left(t^{\frac{4}{3}}+t^{\frac{1}{3}}\right) d t$

$=\frac{1}{3}\left[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]+\mathrm{C}$

$=\frac{1}{3}\left[\frac{3}{7} t^{\frac{7}{3}}+\frac{3}{4} t^{\frac{4}{3}}\right]+\mathrm{C}$

$=\frac{1}{7}\left(x^{3}-1\right)^{\frac{7}{3}}+\frac{1}{4}\left(x^{3}-1\right)^{\frac{4}{3}}+\mathrm{C}$