# One integral, two solutions?

1. Aug 12, 2012

### Cetullah

Hello dear Physics Forums users!

I ve recently passed to 2nd class, however I failed my Math II lesson, so I was solving some problems.

Here is it, with my solution attempt:

∫(x+3)/$\sqrt{}(x^2-4)$

∫x/$\sqrt{}(x^2-4)$ + 3/($\sqrt{}(x^2-4)$

Well eh, screw the integral on left anyway, what really confused me was the one on right:

Here s my solution:

∫3/($\sqrt{}(x^2-4)$=-3∫1/$\sqrt{}(4-x^2)$

=-3arcsin(x/2)

But on the other side, my book and WolframAlpha claims that the solution for the integral on right is:

3 ln(x+$\sqrt{}(x^2-4))$

So I checked what they look like, and here are the results:

http://www.wolframalpha.com/input/?i=∫3/√(x^2-4)

http://www.wolframalpha.com/input/?i=∫-3/√(4-x^2)

So they are TWO DIFFERENT EQUATIONS?

Would my answer be wrong on exam?

Last edited: Aug 12, 2012
2. Aug 12, 2012

### Millennial

You can't move the - inside the radical. To have an - inside the radical means to have the imaginary unit i outside the radical.
Protip: You can use an hyperbolic substitution to evaluate the integral on the right, a trig substitution also works.

3. Aug 12, 2012

### Cetullah

Yeah, that makes quite sense, thanks!

I failed to crush the mathematics again, lawl :)

4. Aug 12, 2012

### GarageDweller

"wtf" and "lame" were tagged for this thread