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Integrate Sqrt[x^2-a]?

  1. Dec 2, 2008 #1
    I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

    does anyone know how to get this/derive this i.e. not take from tables !!!!

  2. jcsd
  3. Dec 2, 2008 #2
    I believe what you can do is write out the expression and take the log of both sides. WHen taking the derivative, you must remember y is a function of x.

    y = Sqrt[x^2 - a]
    y = [x^2 - a]^(1/2)

    Take natural log

    Ln[y] = Ln{ [x^2 - a]^(1/2) }
    Ln[y] = (1/2) Ln{[x^2 - a]}

    Take derivative wrt x. Don't forget you must apply chain rule to right hand side. Where y' comes out anyway.

    (1/y) (y') = 2x/ (2 (x^2 - a))

    writing more neatly, cancel a two

    (y'/y) = x / (x^2 - a)

    Finally, multiply a y back over and resub your origional y.

    y' = yx/ (x^2 - a) y = [x^2 - a]^(1/2)

    y' = x [x^2 - a]^(1/2) / (x^2 - a)

    y' = x(x^2 - a)^(-1/2)
  4. Dec 2, 2008 #3
    I'm an idiot, I took a deravitive. Not even the easiest way either.
  5. Dec 2, 2008 #4


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    Homework Helper

    Use a trigo substitution. Specifically, look at the ones for sec and tan.
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