How does Mathematica get the following answer for the integral in the title
The answer I get is different and I do not understand how can Mathematica have sgrt(1-x^3) in the answer when the original function is sqrt(x^3-1).
Thank you
The answer I get is different
What is the answer you got?
I verified it now by hand, Mathematica is correct. I get the answer in terms of \arctan, little simpler than what Mathematia gives
I=\int\frac{\sqrt{x^{3}-1}}{x}dx
Let u=x^{3}+1, then du=3x^{2}dx and the above becomes
I=\int\frac{\sqrt{u}}{3x^{3}}du=\frac{1}{3}\int\frac{\sqrt{u}}{u-1}dx
Now let u=\tan^{2}v or \sqrt{u}=\tan v, hence $\frac{1}{2}\frac{1}
{\sqrt{u}}du=\sec^{2}vdv$ and the above becomes
\begin{align*}
I & =\frac{1}{3}\int\frac{\sqrt{u}}{\tan^{2}v-1}\left( 2\sqrt{u}\sec
^{2}v\right) dv\
& =\frac{2}{3}\int\frac{u}{\tan^{2}v-1}\sec^{2}vdv\
& =\frac{2}{3}\int\frac{\tan^{2}v}{\tan^{2}v-1}\sec^{2}vdv
\end{align*}
But \tan^{2}v-1=\sec^{2}v hence
\begin{align*}
I & =\frac{2}{3}\int\tan^{2}vdv\
& =\frac{2}{3}\left( \tan v-v\right)
\end{align*}
Substituting back
I=\frac{2}{3}\left( \sqrt{u}-\arctan\left( \sqrt{u}\right) \right)
Substituting back
I=\frac{2}{3}\left( \sqrt{x^{3}+1}-\arctan\left( \sqrt{x^{3}+1}\right)\right)
One can do u- sub u=x^3-1, trig sub x= (sec u)^(2/3), or hyperbolic sub x=(cosh u)^(2/3). The answer is the same.
How can Mathematica be correct when it has hyperbolic tangent and sqrt (1-x^3) where the original function is defined only for x>=1? If you multiply you get
If one uses complex numbers it works since
i arctan(-iz)=arctanh(z).
But my question is why would Mathematica give an answer in complex functions when this is a very simple integral to begin with.
It is because Mathematica was not told to restrict solutions to when x >= 1 and therefore gives a general solution that works with complex values for the function (which in this case happens when x < 1). Therefore the solution might not end up being as simple as expected.
When you do have a function for which you have a restricted range (and especially if outside of that range you might get complex numbers), you can tell Mathematica that in a number of ways (dare I say no pun intended?):
Simplify[Integrate[((x^3 - 1)^(1/2))/x, {x,1,a}],a>1]
which results in
You might sometimes need FullSimplify rather than just Simplify.
This makes sense. Thank you.