# How to solve $\displaystyle\int \bigg(\small\sqrt{\normalsize x+\small\sqrt{\normalsize x+\sqrt{ x+\sqrt{x}}}}\;\normalsize\bigg)\;dx$?

How to solve this:

$$\displaystyle\int \bigg(\small\sqrt{\normalsize x +\small\sqrt{\normalsize x +\small\sqrt{\normalsize x +\sqrt{x}}}}\;\normalsize\bigg) \;dx$$

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Seems like this is a tough one, at least mathematica cannot find it wolframalpha.com/input/… – Listing Nov 24 '11 at 8:07
I don't think that integral is huge enough. – mathmath8128 Nov 24 '11 at 8:07
What does it mean to "solve" an anti-derivative? – Ryan Budney Nov 24 '11 at 8:25
You could try a series expansion. – Raskolnikov Nov 24 '11 at 8:37
I can help with $I=\int \sqrt{x+\sqrt{x}}dx$: Let $u(x)=\sqrt{x}$. Then $dx=2\sqrt{x}du$ and $I$ becomes $I=\int \sqrt{u^2 + u}2udu$. We integrate by parts and $I=2[\frac{u}{2\sqrt{u^2+u}}-\int \sqrt{u^2+u}du]$. Let $J=\int \sqrt{u^2+u}du$. Then $J=\frac{1+2u}{4}\sqrt{u^2+u}-\frac{1}{8}ln |2u+1+2\sqrt{u^2+u}|+c$ (from tables). – Pantelis Sopasakis Nov 24 '11 at 9:53

While this is not an answer, I posted it as an answer so that I can attach the graph. On the graph you can see how close the 2 functions are. Using this fact may help.

The Black graph is the original function and the Red graph is for the approximate function.

A better approximation for the function $f(x)=\displaystyle\bigg(\small\sqrt{\normalsize x+\small\sqrt{\normalsize x+\sqrt{ x+\sqrt{x}}}}\;\normalsize\bigg)$ is:

$0.5(1+\sqrt[]{1+4x})$

This approximation was obtained from Nested Radicals, This formula is an exact value for the infinite case, so it may be used as an approximation only in your case.

A picture of the 2 functions is shown below. The Black graph is the original function and the Red graph is for the approximate function.

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How did you come up with this approximation? Can you post the proof that it asymptotically approximates the given function? – Pantelis Sopasakis Nov 24 '11 at 10:28
In fact I can't claim that it an optimal approximation since I came across it by plotting various curves and noticed that this one was among the better ones. The main point I wanted to show is that there may be a good approximation for the original function that may help. – NoChance Nov 24 '11 at 10:36