Step 1: Differentiate the answer to check
Let $G=\frac{1}{3}\log\left|\frac{u-1}{u+1}\right|$ with $u=\sqrt{1-x^{3}}$.
$\frac{dG}{du}=\frac{1}{3}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)=\frac{2}{3(u^{2}-1)}$.
Step 2: Use du/dx
$\frac{du}{dx}=\frac{-3x^{2}}{2u}$ and $u^{2}-1=-x^{3}$.
Step 3: Multiply
$\frac{dG}{dx}=\frac{2}{3(-x^{3})}\cdot\frac{-3x^{2}}{2u}=\frac{1}{xu}$, which is the integrand.
Step 4: Eliminate
The option with $\frac{2}{3}$ differs by a factor, and the option with the reversed ratio changes the sign, so only the stated form works.
Final Answer:
Hence the integral is one third of the log of the stated ratio plus c.
\[ \boxed{\text{(B) }\dfrac{1}{3}\log \left|\dfrac{\sqrt{1-x^{3}}-1}{\sqrt{1-x^{3}}+1}\right|+c} \]