A Simple integral

1. Jul 4, 2016

felicja

$\int\sqrt{\frac{x}{x-a}}dx=?$

2. Jul 4, 2016

QuantumQuest

What have you tried so far?

3. Jul 4, 2016

4. Jul 4, 2016

micromass

Staff Emeritus
I don't get it. I clicked the wolfram link. What is so strange and ridiculous about the solution?

5. Jul 4, 2016

QuantumQuest

The differences are due to simplifications and / or different ways to write the same thing.
Have you tried to do the integration? I give you the result if you want to try it out:

$\dfrac{a\left(\ln\left(\sqrt{x-a}+\sqrt{x}\right)-\ln\left(\left|\sqrt{x-a}-\sqrt{x}\right|\right)\right)}{2}+\sqrt{x}\sqrt{x-a}$

6. Jul 4, 2016

felicja

I easily compute this integral - just make a simple substitute:
$x = a\cosh^2(t)$
then:
$dx = 2a\cosh(t)\sinh(t)$
and
$x-a = a\sinh^2(t)$
so, the integral now is:
$\int cosh^2(t)dt = \int(\cosh(2t)+1)dt = \frac{1}{2}\sinh(2t)+t+C=\sinh(t)\cosh(t)+t+C$

finally:
$I = \sqrt{x(x-a)}+a\cdot arcosh{\sqrt{x/a}}$

7. Jul 4, 2016

micromass

Staff Emeritus
You need to get back to the variable $x$.

8. Jul 4, 2016

felicja

So, I don't what is going on.

The proposed solutions are quite nonsensical - what it the reason?

$arcosh(x)=\ln(x+\sqrt{x^2-1})$

9. Jul 4, 2016

micromass

Staff Emeritus
Can you tell us why they are nonsensical?

10. Jul 4, 2016

felicja

And there are much more idiotic versions in the net!

11. Jul 4, 2016

micromass

Staff Emeritus
The links you posted in #3 are absolutely correct, so I have no idea what you're talking about.

12. Jul 4, 2016

felicja

Try to compute some definite integral using these 'alternative solutions' then You get it.

For example:
what is a correct distance, means: according to the GR, to the Sun from the Earth?

13. Jul 4, 2016

micromass

Staff Emeritus
Can you please be more specific?

14. Jul 4, 2016

felicja

The question: 'what is a distance...', and with a given metric is rather very precise - there is no room for any more specification.

15. Jul 4, 2016

micromass

Staff Emeritus
You're making no sense, sorry.

16. Jul 6, 2016

felicja

OK, I sorry.
The result is the same.
$2\ln(\sqrt{x}+\sqrt{x-1})=\ln(\sqrt{x}+\sqrt{x-1})^2=\ln(2\sqrt{x}\sqrt{x-1}+x+x-1)$

But look at this.
The distance with the Schwarzschild metric is equal to:
$s=\sqrt{r(r-a)} + a.arcosh(\sqrt{r/a})$
thus for the case of very big distances: $r >> a$, the distance is approximately:
$s\approx r + a\ln(4r/a)$

thus to the Sun it is some bigger distance, than the simple: r = 150 mln km,
because it's about: $ds = a\ln(4r/a)$ bigger.
a = 3km for the Sun, so this is:
$ds = 3\ln(4*150mln / 3) = 36.6 km$ more.

Last edited: Jul 6, 2016