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

Question

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.

Answer 1

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

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

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

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

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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.