# integral problem

• Jul 21st 2010, 11:21 PM
pencil09
integral problem

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

substitution method

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

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

$x+y=y^2$

??????
i've no idea what the next step are.....(Headbang)
• Jul 21st 2010, 11:47 PM
chisigma
Setting...

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

... You derive immediately that is...

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

... so that is...

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

Kind regards

$\chi$ $\sigma$
• Jul 21st 2010, 11:59 PM
pencil09
Quote:

Originally Posted by chisigma
Setting...

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

... You derive immediately that is...

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

... so that is...

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

Kind regards

$\chi$ $\sigma$

i see...

thanks chisigma.... i'll continue from here....
• Jul 22nd 2010, 01:06 AM
chisigma
The question has a little minor problem: setting...

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

... it seems to be that...

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

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

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

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

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

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

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

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

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

... even if I'm not 'cent per cent' sure (Thinking)...

$\chi$ $\sigma$