<s>[INST] Substitution integral with $t=x^3$ 
$$
\int \frac{3}{t(3+\sqrt[3]{t})} \mathrm{d} t . \text { Substitution } t=x^{3}
$$
$$
\int \frac{3}{x^3(3+\sqrt[3]{x^3})}3x^2
 $$
$$
\int \frac{9}{x(3+{x})}
 $$
$$
9\int \frac{1}{3x+{x^2}}
 $$


Now i'm stuck
my calculation so far
 [/INST] $$\int \frac{3}{t(3 + t^{1/3})} \, dt = \int \frac{1}{3 + t^{1/3}} \cdot 9t^{-1/3} \cdot \frac{1}{3} t^{-2/3} \, dt,$$ so with the substitution $$u = 3 + t^{1/3}, \quad du = \frac{1}{3} t^{-2/3} \, dt, \\ t^{-1/3} = \frac{1}{u-3},$$ we obtain $$\int \frac{3}{t(3+t^{1/3})} \, dt = \int \frac{1}{u} \cdot \frac{9}{u-3} \, du = 3 \int \left( \frac{1}{u-3} - \frac{1}{u} \right) \, du = 3 \log\left|\frac{t}{(3+t^{1/3})^3}\right| + C.$$
</s>