$\int \frac{1}{x(1+\sqrt[3]{x})^2} d x$ is equal to

$3\left\{\log \left(\frac{x^{1 / 3}}{1+x^{1 / 3}}\right)+\frac{1}{1+\sqrt[3]{x}}\right\}+C$

Let

$I=\int \frac{1}{x(1+\sqrt[3]{x})^2} d x=\int \frac{1}{t^3(1+t)^2} 3 t^2 d t$

$\Rightarrow I=3 \int \frac{1}{t(t+1)^2} d t=3 \int\left\{\frac{1}{t}-\frac{1}{t+1}-\frac{1}{(t+1)^2}\right\} d t$

$\Rightarrow I=3\left\{\log _e t-\log (t+1)+\frac{1}{t+1}\right\}+C$

$\Rightarrow I=3\left\{\log _e\left(\frac{t}{t+1}\right)+\frac{1}{t+1}\right\}+C$

$\Rightarrow I=3\left\{\log _e\left(\frac{x^{1 / 3}}{1+x^{1 / 3}}\right)+\frac{1}{x^{1 / 3}+1}\right\}+C$