2
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ It's $t^3$ not $t$ at denominator $\endgroup$
    – Eureka
    Mar 27 '19 at 16:50
  • 1
    $\begingroup$ Why there is t in the denominator ($t^3=x$) . It should be $t^3$ $\endgroup$
    – Tojrah
    Mar 27 '19 at 16:53
  • 1
    $\begingroup$ Mathematica: $6 \sqrt{\sqrt[3]{x}-2}-6 \sqrt{2} \tan ^{-1}\left(\frac{\sqrt{\sqrt[3]{x}-2}}{\sqrt{2}}\right)$ $\endgroup$ Mar 27 '19 at 16:57
2
$\begingroup$

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

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

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

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.

$\endgroup$
4
  • $\begingroup$ I would follow up with the trigonometric substitution. $$t=2\sec^2 u, dt = 4\sec^2 u \tan u$$ so that gives $$6\sqrt{2}\int{\tan^2 u du}$$ $\endgroup$ Mar 27 '19 at 17:11
  • $\begingroup$ @interstellarprobe Are you sure? There's no $t^2$ under the square root, just a $t$... $\endgroup$
    – StackTD
    Mar 27 '19 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. $\endgroup$ Mar 27 '19 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 '19 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.