If int frac{dx}{x sqrt{1-x^3}} = k log left(f | vedclass

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

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$\frac{1}{3}$

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(C) Let $I = \int \frac{dx}{x \sqrt{1-x^3}}$.Multiply the numerator and denominator by $x^2$:$I = \int \frac{x^2 dx}{x^3 \sqrt{1-x^3}}$.Let $1-x^3 = t^2$,then $-3x^2 dx = 2t dt$,so $x^2 dx = -\frac{2}{3} t dt$.Also,$x^3 = 1-t^2$.Substituting these into the integral:$I = \int \frac{-\frac{2}{3} t dt}{(1-t^2) t} = -\frac{2}{3} \int \frac{dt}{1-t^2} = \frac{2}{3} \int \frac{dt}{t^2-1}$.Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}ight| + c$:$I = \frac{2}{3} 	imes \frac{1}{2} \log \left|\frac{t-1}{t+1}ight| + c = \frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}ight| + c$.Comparing this with the given expression,we find $k = \frac{1}{3}$.

If $\int \frac{dx}{x \sqrt{1-x^3}} = k \log \left(\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right) + c$,(where $c$ is a constant of integration),then the value of $k$ is

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