If $\int \frac{1}{x\sqrt{1-x^3}}dx=\frac{a}{9}\log \left|\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}+1}\right|+b,$ then $a$ is equal to

Solution:

Putting $1-x^3=y^2,-3x^{2}dx=2ydy$, we get

$I=\int \frac{1}{x\sqrt{1-x^3}}dx=-\frac{2}{3}\int \frac{1}{1-y^2}dy=\frac{2}{3}\int \frac{1}{y^2-1}dy$

$\Rightarrow I=\frac{1}{3}\log \left|\frac{y-1}{y+1}\right|+C=\frac{1}{3}\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$

Hence, $a=3$