2
$\begingroup$

How to solve this:

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

  • 2
    Seems like this is a tough one, at least mathematica cannot find it http://www.wolframalpha.com/input/?i=Integrate[Sqrt[x+%2B+Sqrt[x+%2B+Sqrt[x%2BSqrt[x]]]]%2Cx]2011-11-24
  • 3
    I don't think that integral is huge enough.2011-11-24
  • 2
    What does it mean to "solve" an anti-derivative?2011-11-24
  • 0
    You could try a series expansion.2011-11-24
  • 2
    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).2011-11-24
  • 0
    Just let $u=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}$ is OK. Then you will get $x$ in terms of polynomial of $u$. WolframAlpha canot solve it because the process time is excessed.2012-07-13

1 Answers 1