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

\(\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{x \sqrt{1-x^3}} \\ & =\int \frac{x^2 \mathrm{~d} x}{x^3 \sqrt{1-x^3}} \mathrm{~d} x \\ & \text { Put } 1-x^3=\mathrm{t}^2 \\ & \Rightarrow-3 x^2 \mathrm{~d} x=2 \mathrm{tdt} \\ & \Rightarrow x^2 \mathrm{~d} x=\frac{-2 \mathrm{tdt}}{3} \\ & \end{aligned}\)
\(\begin{aligned} \therefore \quad \mathrm{I} & =\int \frac{-\left(\frac{2 \mathrm{tdt}}{3}\right)}{\left(1-\mathrm{t}^2\right) \mathrm{t}} \\ & =-\frac{2}{3} \int \frac{1}{1-\mathrm{t}^2} \mathrm{dt} \\ & =\frac{2}{3} \int \frac{1}{\mathrm{t}^2-1} \mathrm{dt} \\ & =\frac{2}{3} \times \frac{1}{2} \cdot \log \left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|+\mathrm{c} \\ & =\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+\mathrm{c} \\ \therefore \quad \mathrm{k} & =\frac{1}{3}\end{aligned}\)