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How do I integrate this? x/(x^2-1)^.5

  1. Nov 16, 2009 #1
    How do I integrate this? 1/(x^2-1)^.5

    How do I integrate this? x/(x^2-1)^.5

    And this
    1/(x^2-1)^.5
     
    Last edited: Nov 16, 2009
  2. jcsd
  3. Nov 16, 2009 #2
    For the first, make a u-substitution.

    For the second, make a trigonometric substitution suggested by that difference of squares.
     
  4. Nov 16, 2009 #3
    Thanks!

    First one worked like a charm.

    For second one I substituted x for cosh(y). Since cosh(y)^2-1=sinh(y)^2, but bottom turns into sinh(y). And since x=cosh(y), dx/dy= sinh(y).

    Back to original equation:
    integral( 1/(x^2-1)^.5 dx) = integral ( sinh(y)/sinh(y)) dy = 1 + constant. However, the answer is supposed to be cosh^-1(x).
     
  5. Nov 16, 2009 #4
    Let x=secø
    dx=secøtanødø
    tanø=√(x2-1)

    So your integral becomes:
    ∫dx/√(x2-1) = ∫secødø = ln|secø+tanø|

    Substituting back in
    ln|secø+tanø|=ln|x+√(x2-1)|
     
  6. Nov 17, 2009 #5
    Think about that some more
     
  7. Nov 17, 2009 #6
    By u substitution, let u=x^2-1 then du=2x
    you have x you only need 2
     
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