Integrate the function $\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}$.

Let $x^{3}-1=t$.
Then,$3x^{2}dx = dt$,which implies $x^{2}dx = \frac{dt}{3}$.
We can rewrite the integral as:
$\int \left(x^{3}-1\right)^{\frac{1}{3}} x^{5} dx = \int \left(x^{3}-1\right)^{\frac{1}{3}} x^{3} \cdot x^{2} dx$.
Substituting $x^{3} = t+1$ and $x^{2}dx = \frac{dt}{3}$,we get:
$= \int t^{\frac{1}{3}}(t+1) \frac{dt}{3} = \frac{1}{3} \int \left(t^{\frac{4}{3}} + t^{\frac{1}{3}}\right) dt$.
Integrating term by term:
$= \frac{1}{3} \left[ \frac{t^{\frac{7}{3}}}{\frac{7}{3}} + \frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right] + C = \frac{1}{3} \left[ \frac{3}{7} t^{\frac{7}{3}} + \frac{3}{4} t^{\frac{4}{3}} \right] + C$.
$= \frac{1}{7} t^{\frac{7}{3}} + \frac{1}{4} t^{\frac{4}{3}} + C$.
Substituting back $t = x^{3}-1$:
$= \frac{1}{7} \left(x^{3}-1\right)^{\frac{7}{3}} + \frac{1}{4} \left(x^{3}-1\right)^{\frac{4}{3}} + C$,where $C$ is an arbitrary constant.