We have $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{A} \log \left(\frac{\sqrt{1-\mathrm{x}^3}+\mathrm{B}}{\sqrt{1-\mathrm{x}^3}+1}\right)\right)=\frac{1}{\mathrm{x} \sqrt{1-\mathrm{x}^3}}$
now integrating both sides
$$
\begin{aligned}
& \int \frac{1}{x \sqrt{1-x^3}} d x=A \log \left(\frac{\sqrt{1-x^3}+B}{\sqrt{1-2^3}+1}\right)+c \\
& =\int \frac{3 x^2 d x}{x^3 \sqrt{1-x^3}} \\
& \text { put } \sqrt{1-x^3}=t \Rightarrow x^3=1-t^2 \\
& \Rightarrow \frac{1}{2}\left(1-x^3\right)^{-\frac{1}{2}}\left(-3 x^2\right) d x=d t \\
& \Rightarrow \frac{-3}{2} \frac{x^2 d x}{\sqrt{1-x^3}}=d t \\
& \frac{1}{3} \int \frac{2}{t^2-1} d t \\
& =\frac{1}{3} \int \frac{(t+1)-(t-1)}{(t+1)(t-1)} d t \\
& =\frac{1}{3} \int \frac{1}{(t-1)}-\frac{1}{(t+1)} d t \\
& =\frac{1}{3}\left[\frac{\log \sqrt{1-x^3}-1}{\log \sqrt{1+x^3+1}}\right]+c \\
& \Rightarrow \frac{1}{3}, B=-1
\end{aligned}
$$