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integral ( (x^3-1)^(1/2)/x )

Posted 10 years ago

How does Mathematica get the following answer for the integral in the title enter image description here

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

POSTED BY: aida galeb
5 Replies
Posted 10 years ago

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

Integral

You might sometimes need FullSimplify rather than just Simplify.

POSTED BY: Jim Baldwin

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

$$ I=\frac{2}{3}\left( \sqrt{x^{3}+1}-\arctan\left( \sqrt{x^{3}+1}\right)\right) $$

POSTED BY: Nasser M. Abbasi
Posted 10 years ago

enter image description here

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 enter image description here

POSTED BY: aida galeb
Posted 10 years ago

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.

POSTED BY: aida galeb
Posted 10 years ago

This makes sense. Thank you.

POSTED BY: aida galeb
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