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.

You have found the correct answer already. Both solutions are the same, they are just written in a different fashion.
If you use the property of the logarithm you can rewrite your initial results by 
$$
\frac{2}{3} \log |\cdots| = \frac{1}{3} \log |\cdots|^2 = \frac{1}{3} \log \left|\frac{2 - x^3 - 2 \sqrt{1-x^3}}{x^3} \right|
$$
In the case of the given answer you can rewrite it by
$$
\frac{1}{3} \log \left| \frac{\sqrt{1-x^3} -1}{\sqrt{1-x^3} +1} \cdot\frac{\sqrt{1-x^3} -1}{\sqrt{1-x^3} -1}\right|
$$
and work out numerator and denominator. This gives the same expression.
Normally you would need to be a bit careful with the sign of the argument in the logarithm, but here with the absolute values that is automatically taken care of.
