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.
 A: 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
$$I=\frac {1}{3}\int \frac {2du}{u^2-1} $$
$$=\frac {1}{3}\int (\frac {1}{u-1}-\frac {1}{u+1})du $$
A: 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.
A: 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.
A: $u=x^3, du=3x^2dx, \frac{dx}{x\sqrt{1-x^3}}=\frac{du}{3u\sqrt{1-u}}$, set here $t=\sqrt{1-u}, u=1-t^2, du=-2t dt$, $\frac{du}{3u\sqrt{1-u}}=\frac{-2dt}{3(1-t^2)}=\frac{1}{3}\left(\ln(1-t)-\ln(1+t)\right)+C$
A: \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}
