If $\int \frac{1}{x \sqrt{1-x^3}} d x=a \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+b$ then a is equal to

$\frac{1}{3}$

Multiplying above and below by x² and put

$1-x^3=t^2$     ∴ $-3 x^2 dx=2 t~dt$

∴  $I=\frac{2}{3} \int \frac{d t}{t^2-1}=\frac{2}{3} . \frac{1}{2} \log \frac{t-1}{t+1}+c$

$I=\frac{1}{3} \log \frac{t-1}{t+1}+c$    ∴ $a=\frac{1}{3}$

Hence (2) is the correct answer.