# Calculating $\int \frac{\sqrt{\sqrt{x} - 2}}{x}dx$

Can someone help me calculate this integral?

$$\int \frac{\sqrt{\sqrt{x} - 2}}{x}dx$$

I tried this substitution:

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

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:

$$t = \sqrt{x} - 2$$

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

But this seems algrebraically impossible to me.

Help's appreciated.

• It's $t^3$ not $t$ at denominator Mar 27, 2019 at 16:50
• Why there is t in the denominator ($t^3=x$) . It should be $t^3$ Mar 27, 2019 at 16:53
• Mathematica: $6 \sqrt{\sqrt{x}-2}-6 \sqrt{2} \tan ^{-1}\left(\frac{\sqrt{\sqrt{x}-2}}{\sqrt{2}}\right)$ Mar 27, 2019 at 16:57

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

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

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

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.

• I would follow up with the trigonometric substitution. $$t=2\sec^2 u, dt = 4\sec^2 u \tan u$$ so that gives $$6\sqrt{2}\int{\tan^2 u du}$$ Mar 27, 2019 at 17:11
• @interstellarprobe Are you sure? There's no $t^2$ under the square root, just a $t$... Mar 27, 2019 at 17:15
• 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. Mar 27, 2019 at 18:02
• I missed the square, I was en route and reading on my smartphone :o). Yes sure, that works too! Mar 27, 2019 at 18:49