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An intergration

  1. Mar 28, 2005 #1
    Does anyone know how to intergrate [tex] \frac{1}{\sqrt{2\beta x-\alpha x^2}}[/tex]
    I went to wolfram and type it in, but it gave me a weird number.
     
  2. jcsd
  3. Mar 28, 2005 #2

    Zurtex

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  4. Mar 28, 2005 #3

    dextercioby

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    It can't give u a #,but a function of "x"...

    Daniel.
     
  5. Mar 28, 2005 #4
    But the answer contain i, so I don't know it that's correct or not
    The last time I put in 1/Sqrt[a x^2], it gave me [x log x]/Sqrt[a x^2]
    Which I don't think is the right answer
     
    Last edited: Mar 28, 2005
  6. Mar 28, 2005 #5
    If you aren't sure whether an indefinite integral is right or not, just differentiate it! If it is then you'll get back your original function (for example, the answer it gave you in your last post is right).
     
  7. Mar 28, 2005 #6
    According to Wolfram's Integrator, [tex]\int 0 dx[/tex] is 0. I thought it was to be a constant.
     
  8. Mar 28, 2005 #7
    I thought the intergral of [tex]\frac{1}{\sqrt{a x^2}}[/tex] is [tex]\frac{\ln{x}}{\sqrt{a}}[/tex]
    Also when I substitute 2b as 59, the integrator gave me an different answer
     
    Last edited: Mar 28, 2005
  9. Mar 28, 2005 #8

    Zurtex

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    It never adds the constant.

    [tex]\frac{x \log x}{\sqrt{a x^2}} = \frac{x \log x}{x \sqrt{a}}[/tex]

    As it goes that web site does have bug I've found but I really doubt you will find them.
     
  10. Mar 28, 2005 #9
    Ok, another question when I put in 1/Sqrt[19.66 x-a x^2], I don't know what "0." in that intergral means. Also it has 1. ax and 2. Sqrt[x]
    what does 0. 1. 2. means?
     
  11. Mar 29, 2005 #10

    dextercioby

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    These integrals are simple enough to them by hand.There's no need to use software to do them for you.The integrator from wolfram is excellent,though i've caught him with some functions he woudn't integrate.

    Okay,there's a problem with your integrals,though.U have a possibly negative expression under a radical.That's why the result is weird,because it may contain complex functions,though the input is real.I suggest u decide on which intervals u wish to integrate...

    Daniel.
     
  12. Mar 29, 2005 #11

    saltydog

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    Integrands like these require you to complete the square inside the radical. You'll then get something like:

    [tex]\sqrt{a^2-(x-b)^2}[/tex]

    in the denominator. In this particular form, the answer can be expressed in terms of the ArcSin function.
     
  13. Mar 29, 2005 #12

    dextercioby

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    Integrals like these

    [tex] \int \frac{dx}{\sqrt{ax^{2}+bx+c}},\int \sqrt{ax^{2}+bx+c} \ dx [/tex]

    are always expressible through elementary functions...

    While even this one

    [tex] \int \sqrt{x^{3}+0,0000000000000000000000000000000000000000001} \ dx [/tex]

    cannot;


    Daniel.
     
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