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

To finish your calculation:
$$\int\frac{9}{x(3+x)}$$
Take partial fractions
$$=3\int \frac{1}{x+3}dx+2\int\frac{1}{x}dx$$
Let $u=x+3,du=dx$ have
$$=3\int\frac{1}{u}du+2\int\frac{1}{x}dx$$
Then exists $C\in\mathbb{R},s.t.$
$$=3\ln(u)+2\ln(x)+C$$
Substitute $u$ back, we have
$$=3\ln(x+3)+2\ln(x)+C$$
