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.

\begin{align}
u & = \sqrt{1-x^3} \\
u^3 & = 1-x^3 \\
3u^2\,du & = -3x^2\,dx \\
u^2\,du & = -x^2 \, dx \\[10pt]
\int \frac{dx}{x\sqrt{1-x^3}} & = \int \frac{x^2\,dx}{x^3\sqrt{1-x^3}} = -\int \frac{u^2\,du}{(1-u^3) u} = -\int \frac{u\,du}{1-u^3} \\[10pt]
& = - \int \frac{u\,du}{(1-u)(1+u+u^2)} = \int \left( \frac A {1-u} + \frac{Bu+C}{1+u+u^2} \right) du \quad \text{etc.}
\end{align}
$$ {} $$
\begin{align}
x^3 & = \sin^2\theta \\
3x^2\,dx & = 2\sin\theta\cos\theta\,d\theta \\[10pt]
\int\frac{dx}{x\sqrt{1-x^3}} & = \int \frac{x^2\,dx}{x^3\sqrt{1-x^3}} = \frac 2 3 \int \frac{\sin\theta\cos\theta\,d\theta}{\sin^2\theta \cos\theta} = \frac 2 3 \int \frac{d\theta}{\sin\theta} \\[10pt]
& = -\frac 2 3 \log|\csc\theta+\cot\theta| + C = - \frac 2 3 \log\left| \frac{1+\cos\theta}{\sin\theta} \right| + C \\
& = -\frac 2 3 \log\left| \frac{1 + \sqrt{1-x^3}}{x^{3/2}} \right| + C
\end{align}
