1. ## integral problem

$\displaystyle \int(x+(x+(x+....)^\frac{1}{2})^\frac{1}{2})^\frac {1}{2}dx=....$

substitution method

$\displaystyle (x+(x+(x+....)^\frac{1}{2})^\frac{1}{2})^\frac{1}{ 2}=y$

$\displaystyle x+(x+(x+....)^\frac{1}{2})^\frac{1}{2}=y^2$

$\displaystyle x+y=y^2$

??????
i've no idea what the next step are.....

2. Setting...

$\displaystyle \displaystyle y= \sqrt {x + \sqrt{x + \sqrt{x + ...}}}$ (1)

... You derive immediately that is...

$\displaystyle \displaystyle \sqrt{x + y} = y \rightarrow y = \frac{1 + \sqrt{1+4x}}{2}$ (2)

... so that is...

$\displaystyle \displaystyle \int \sqrt {x + \sqrt{x + \sqrt{x + ...}}}\ dx = \int \frac{1 + \sqrt{1+4x}}{2}\ dx$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
Setting...

$\displaystyle \displaystyle y= \sqrt {x + \sqrt{x + \sqrt{x + ...}}}$ (1)

... You derive immediately that is...

$\displaystyle \displaystyle \sqrt{x + y} = y \rightarrow y = \frac{1 - \sqrt{1+4x}}{2}$ (2)

... so that is...

$\displaystyle \displaystyle \int \sqrt {x + \sqrt{x + \sqrt{x + ...}}}\ dx = \int \frac{1 - \sqrt{1+4x}}{2}\ dx$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
i see...

thanks chisigma.... i'll continue from here....

4. The question has a little minor problem: setting...

$\displaystyle \displaystyle y(x) = \sqrt {x + \sqrt{x + \sqrt{x + ...}}}$ (1)

... it seems to be that...

a) is $\displaystyle y(0)=0$ ...

b) $\displaystyle y(x)$ is increasing with x...

Now is a little difficult to extablish which of the two solutions of the equation...

$\displaystyle y^{2} - y - x =0$ (2)

... meets both the requirements a) and b), because...

c) $\displaystyle \displaystyle y(x)= \frac{1 - \sqrt{1+4x}}{2}$ vanishes in $\displaystyle x=0$ but is decreasing with x...

d) $\displaystyle \displaystyle y(x)= \frac{1 + \sqrt{1+4x}}{2}$ is increasing with x but is $\displaystyle y(0)=1$ ...

Since it is requested fo find a primitive of (1) and all primitives are defined unless a constant, in my opinion the solution is...

$\displaystyle \displaystyle \int \sqrt {x + \sqrt{x + \sqrt{x + ...}}}\ dx = \int \frac{1 + \sqrt{1+4x}}{2}\ dx$ (3)

... even if I'm not 'cent per cent' sure ...

$\displaystyle \chi$ $\displaystyle \sigma$