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

You could do partial fraction decomposition, but instead let's multiply the top and bottom by $\frac{1}{x^2}$:
$$3\int\frac{3}{3x+x^2}dx = 3\int \frac{\frac{3}{x^2}}{\frac{3}{x}+1}dx$$
Notice that the numerator is the derivative of the denominator times $-1$, so we have an integral of the form $\int\frac{du}{u}$, which integrates to
$$ -3\log\left(\frac{3}{x}+1\right)+C$$
Then substitute back in for $t$:
$$ -3\log\left(\frac{3}{\sqrt[3]{t}}+1\right)+C$$
