Correct answer of an indefinite integral Find the value of 
$$ \int{\frac{dx}{x\sqrt{1-x^3}}} $$
I assumed $x^3 = \sin^2\theta$ and found the solution as
$$\frac{2}{3} \log\left|\frac{1}{x\sqrt{x}} - \frac{\sqrt{1-x^3}}{x\sqrt{x}} \right| + c$$
but the solution is given as
$$\frac{1}{3} \log{\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|} + c$$
Any help to reach to this provided solution will be appreciated.

Please multiply numerator and denominator by $x^2$ and then put $t=x^3$ with $dt=3x^2dx $.
you will get $$I=\frac {1}{3}\int \frac {dt}{t\sqrt {1-t}} $$
to finish, put $$u=\sqrt {1-t }.$$
to find
$$I=\frac {1}{3}\int \frac {2du}{u^2-1} $$
$$=\frac {1}{3}\int (\frac {1}{u-1}-\frac {1}{u+1})du $$
