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.

Note that
$$
\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}=
\frac{(\sqrt{1-x^3}-1)^2}{(\sqrt{1-x^3}+1)(\sqrt{1-x^3}-1)}
=
-\frac{(\sqrt{1-x^3}-1)^2}{x^3}.
$$
Hence
$$
\frac{1}{3}\ln\biggl|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\biggr|
=
\frac{2}{3}\ln\biggl|\frac{1-\sqrt{1-x^3}}{x^{3/2}}\biggr|.
$$
Thus, your answer is also correct.
