If ∫1x1−x3dx=a9log⁡1−x2−11−x2+1+b, then a is equal to

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

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

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

Hence, $a = 3$

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