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One integral, two solutions?

  1. Aug 12, 2012 #1
    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=∫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?

    Thanks for your help!
     
    Last edited: Aug 12, 2012
  2. jcsd
  3. Aug 12, 2012 #2
    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.
     
  4. Aug 12, 2012 #3
    Yeah, that makes quite sense, thanks!

    I failed to crush the mathematics again, lawl :)
     
  5. Aug 12, 2012 #4
    "wtf" and "lame" were tagged for this thread
     
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