Can someone help me calculate this integral?

$\int \frac{\sqrt{\sqrt[3]{x} - 2}}{x}dx$

I tried this substitution:

$\Bigg(t = \sqrt[3]{x}, t^3 = x, 3t^2dt=dx\Bigg)$

which reduces the integral to:

$\int \frac{\sqrt{t-2}}{t}3t^2dt = 3\int \sqrt{t-2}tdt$

and continuing from here is pointless, because the result (according to wolfram) is waaay wrong. I don't understand why that substitution was wrong...

So I also tried this substitution instead:

$t = \sqrt[3]{x} - 2$

$t^3 = x - 6 \sqrt[3]{x^2} + 12\sqrt[3]{x} - 8$

But this seems algrebraically impossible to me.

Help's appreciated.

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    $\begingroup$ It's $t^3$ not $t$ at denominator $\endgroup$ – Eureka Mar 27 at 16:50
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    $\begingroup$ Why there is t in the denominator ($t^3=x$) . It should be $t^3$ $\endgroup$ – Tojrah Mar 27 at 16:53
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    $\begingroup$ Mathematica: $6 \sqrt{\sqrt[3]{x}-2}-6 \sqrt{2} \tan ^{-1}\left(\frac{\sqrt{\sqrt[3]{x}-2}}{\sqrt{2}}\right)$ $\endgroup$ – David G. Stork Mar 27 at 16:57

$\int \frac{\sqrt{\sqrt[3]{x} - 2}}{\color{blue}{x}}dx$

$\Bigg(t = \sqrt[3]{x}, \color{blue}{t^3 = x}, 3t^2dt=dx\Bigg)$

$\int \frac{\sqrt{t-2}}{\color{red}{t}}3t^2dt = \ldots$

With your substitution, you want $\color{blue}{t^3}$ where you have $\color{red}{t}$.

You then end up with: $$3\int\frac{\sqrt{t-2}}{t}\,\mbox{d}t$$ and you can follow up with e.g. $u=\sqrt{t-2}$ to rationalize the integrand.

  • $\begingroup$ I would follow up with the trigonometric substitution. $$t=2\sec^2 u, dt = 4\sec^2 u \tan u$$ so that gives $$6\sqrt{2}\int{\tan^2 u du}$$ $\endgroup$ – InterstellarProbe Mar 27 at 17:11
  • $\begingroup$ @interstellarprobe Are you sure? There's no $t^2$ under the square root, just a $t$... $\endgroup$ – StackTD Mar 27 at 17:15
  • $\begingroup$ quite sure. $$\sqrt{t-2} = \sqrt{2\sec^2 u -2} = \sqrt{2}\tan u$$ the denominator becomes $$2\sec^2 u$$ and $$dt = 4\sec^2 u \tan u$$ It all works out great. $\endgroup$ – InterstellarProbe Mar 27 at 18:02
  • $\begingroup$ I missed the square, I was en route and reading on my smartphone :o). Yes sure, that works too! $\endgroup$ – StackTD Mar 27 at 18:49

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