If frac{d}{dx} left( A log left( frac{sqrt{1- | vedclass

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(B) Let $I = \int \frac{1}{x \sqrt{1-x^3}} dx$. Multiply numerator and denominator by $x^2$: $I = \int \frac{x^2}{x^3 \sqrt{1-x^3}} dx$. Let $t = \sqr

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$\frac{-1}{3}$

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(B) Let $I = \int \frac{1}{x \sqrt{1-x^3}} dx$. Multiply numerator and denominator by $x^2$: $I = \int \frac{x^2}{x^3 \sqrt{1-x^3}} dx$. Let $t = \sqrt{1-x^3}$,then $t^2 = 1-x^3$,so $x^3 = 1-t^2$. Differentiating,$2t dt = -3x^2 dx$,so $x^2 dx = -\frac{2}{3} t dt$. Substituting these into the integral: $I = \int \frac{-\frac{2}{3} t dt}{(1-t^2) t} = -\frac{2}{3} \int \frac{dt}{1-t^2} = \frac{2}{3} \int \frac{dt}{t^2-1} = \frac{2}{3} \cdot \frac{1}{2} \log \left| \frac{t-1}{t+1} ight| + C = \frac{1}{3} \log \left| \frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1} ight| + C$. Comparing this with the given expression $A \log \left( \frac{\sqrt{1-x^3}+B}{\sqrt{1-x^3}+1} ight)$,we get $A = \frac{1}{3}$ and $B = -1$. Therefore,$AB = \frac{1}{3} 	imes (-1) = -\frac{1}{3}$.

If $\frac{d}{dx} \left( A \log \left( \frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1} \right) \right) = \frac{1}{x \sqrt{1-x^3}}$,then $AB=$

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