If $\int \frac{dx}{x\sqrt{1-x^3}}=a\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+c$ , then

${\displaystyle \int \frac{dx}{x\sqrt{1-x^3}}}={\displaystyle \int \frac{x^{2}dx}{x^3\sqrt{1-x^3}}}$  let $1-x^3=t^2,-3x^{2}dx=2tdt$ 

$I=-\frac{2}{3}{\displaystyle \int \frac{tdt}{\left(1-t^2\right)t}=-\frac{2}{3}{\displaystyle \int \frac{dt}{1-t^2}=\frac{2}{3}{\displaystyle \int \frac{dt}{t^2-1}}}}$

$=\frac{1}{3}\log \left|\frac{t-1}{t+1}\right|+c=\frac{1}{3}\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+c$ $\Rightarrow a=\frac{1}{3}$ .