Step 1: Plan - integrate $ \frac{1}{x\sqrt{1-x^3}} $ and match.
Instead of differentiating the given form, we find $ \int\frac{dx}{x\sqrt{1-x^3}} $ and identify $ A $ and $ B $ by comparison.
Step 2: Substitute $ u=\sqrt{1-x^3} $.
$ u^2=1-x^3 $, $ x^3=1-u^2 $, $ 2u\,du=-3x^2\,dx $.
Step 3: Transform the integral.
\[ \int\frac{dx}{x\cdot u} = \int\frac{-2u\,du}{3x^3\cdot u} = \frac{-2}{3}\int\frac{du}{1-u^2} \] using $ x^3=1-u^2 $.
Step 4: Apply partial fractions: $ \frac{1}{1-u^2}=\frac{1/2}{1-u}+\frac{1/2}{1+u} $.
\[ I = \frac{-1}{3}\int\!\left(\frac{1}{1-u}+\frac{1}{1+u}\right)du = \frac{1}{3}\ln|1-u|-\frac{1}{3}\ln|1+u|+C = \frac{1}{3}\ln\frac{1-u}{1+u}+C \]
Step 5: Substitute back $ u=\sqrt{1-x^3} $ and match with the given form.
\[ I = \frac{1}{3}\ln\frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}+C \] Comparing with $ A\log\frac{\sqrt{1-x^3}+B}{\sqrt{1-x^3}+1} $: we see $ A=\frac{1}{3} $ and $ B=-1 $ (after rewriting $ \frac{1-u}{1+u}=\frac{-(u-1)}{u+1}=\frac{u+(-1)}{u+1}\cdot(-1) $, absorbing the sign into $ A $ gives $ A=-\frac{1}{3} $, $ B=1 $... either way $ AB=-\frac{1}{3} $).
Step 6: State AB.
\[ AB = \frac{1}{3}\times(-1) = -\frac{1}{3} \] \[ \boxed{-\dfrac{1}{3}} \]