$\int \frac{1}{x \sqrt{1-x^3}} d x$ is equal to

$\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$

We have,

$I=\int \frac{1}{x \sqrt{1-x^3}} d x$

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

$\Rightarrow I =-\frac{1}{3} \int \frac{1}{x^3 \sqrt{1-x^3}} d\left(1-x^3\right)$

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

$\Rightarrow I=-\frac{2}{3} \int \frac{1}{1-t^2} d t=\frac{2}{3} \int \frac{1}{t^2-1^2} d t=\frac{1}{3} \log \left|\frac{t-1}{t+1}\right|+C$

$\Rightarrow I=\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$