One integral, two solutions?

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)/[itex]\sqrt{}(x^2-4)[/itex]

∫x/[itex]\sqrt{}(x^2-4)[/itex] + 3/([itex]\sqrt{}(x^2-4)[/itex]

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

Here s my solution:

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

=-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+[itex]\sqrt{}(x^2-4))[/itex]

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

http://www.wolframalpha.com/input/?i...A%28x%5E2-4%29

http://www.wolframalpha.com/input/?i...A%284-x%5E2%29

So they are TWO DIFFERENT EQUATIONS?

Would my answer be wrong on exam?

Thanks for your help!

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.

Cetullah

Yeah, that makes quite sense, thanks!

I failed to crush the mathematics again, lawl :)

GarageDweller

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