Calculus Challenge

Let's see who's got the brain to solve this problem.

Compute the integral

[br]\displaystyle I= \int \sqrt{2+\sqrt{2+...+\sqrt{2+x}}}dx[br]

where the expression in the integrand contains

n\geq 1

square roots.

Let's see who's got the brain to solve this problem.

THe integral may be written form bellow

\displaystyle \int \sqrt{\sqrt{\sqrt{....\sqrt{x+2}+2}+2}+2}\ dx

And can be solved with series of substitution
First:
a=x+2 and dx=da
so

\displaystyle \int \sqrt{\sqrt{....\sqrt{\sqrt{a}+2}+2}+2}\ da

now substitute
b=Va ->a=b^2 ->da=2b db so

\displaystyle 2\int b\sqrt{\sqrt{....\sqrt{b+2}+2}+2}\ db

substitute

c=\sqrt{b+2}

-> b=c^2-2 ->db=2c dc
so

\displaystyle 4\int c\cdot (c^2-2)\sqrt{\sqrt{....\sqrt{c+2}+2}+2}\ dc

and continue with similar substitution up to there is not sqare root.
You will for example substituting n-times (n square root) and at last with z

\displaystyle 2^n\int z(z^2-2)((z^2-2)^2-2)\cdot ..... \cdot (((..(z^2-2)^2....)^2-2)^2-2)\ dz

Then multplying the powers integrate by terms and resubstitute.
It means long and systematic work.
For an example:
http://www.wolframalpha.com/input/?i=integrate+sqrt%28+2%2Bsqrt%282%2B+sqrt%282%2Bsqrt%282%2Bx%29%29%29%29+dx

(edited 13 years ago)

Let's see who's got the brain to solve this problem.

It would be better, If you post this in here

see if you can get an ODE out of that by differentiating and squaring

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