<s>[INST] 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.
 [/INST] $u=x^3, du=3x^2dx, \frac{dx}{x\sqrt{1-x^3}}=\frac{du}{3u\sqrt{1-u}}$, set here $t=\sqrt{1-u}, u=1-t^2, du=-2t dt$, $\frac{du}{3u\sqrt{1-u}}=\frac{-2dt}{3(1-t^2)}=\frac{1}{3}\left(\ln(1-t)-\ln(1+t)\right)+C$
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