Question

If $\int \frac{1}{x\sqrt{1-x^3}}dx=a\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+b,$ then $6a$ is equal to
$(\log \text{is taken on natural base}{}^{\prime }{e}^{\prime }$ and $C$ is constant of integration)

Answer 1

Putting $1-x^3=y^2$
$\Rightarrow -3x^{2}dx=2ydy,$ we get
$\int \frac{1}{x\sqrt{1-x^3}}dx=-\frac{2}{3}\int \frac{1}{1-y^2}dy=\frac{1}{3}\log \left|\frac{y-1}{y+1}\right|+C=\frac{1}{3}\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C\Rightarrow a=\frac{1}{3}$