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Integral with radical denominator.

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data


    Need to do this integral as part of a problem

    [itex]\int[/itex] [itex]\frac{xdx}{\sqrt{x^2+a^2-\sqrt{2}ax}}[/itex]


    I can't think of any substitution that would work, or any way to factorize.
    I think this is a standard integral, but can't remember everyone so I just want to learn to solve it.
     
  2. jcsd
  3. Nov 19, 2011 #2

    Dick

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    Your first step would be completing the square in the denominator. What does the integral look like after you do that?
     
  4. Nov 19, 2011 #3

    LCKurtz

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    I would try writing [itex]x^2+a^2-\sqrt 2 ax = x^2- \sqrt 2 ax+ \frac{a^2} 2 +(a^2-\frac {a^2} 2)
    =(x-\frac a {\sqrt2})^2+\frac{a^2} 2[/itex] and see if that leads anywhere.

    [Edit] I see Dick types faster than I do...
    [Edit] Fixed a typo and its ramifications.
     
    Last edited: Nov 19, 2011
  5. Nov 19, 2011 #4

    Dick

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    No, we type about the same. I just say less.
     
  6. Nov 19, 2011 #5

    SammyS

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    It appears that you have changed [itex]\sqrt{2}ax[/itex] into [itex]2\sqrt{a}[/itex], but apparently meant [itex]2(\sqrt{a})x\,.[/itex]
     
  7. Nov 19, 2011 #6

    LCKurtz

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    Yes, thanks. I edited it (about 3 times :frown:) to correct it.
     
  8. Nov 19, 2011 #7

    SteamKing

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    You are letting the radicals and the denominator distract you. If you re-write as
    Int (mess)^(-1/2) * xdx, you should see a u-substitution and integration by parts.
     
  9. Nov 23, 2011 #8
    ok when I complete the squares and do it, i get the answer as


    [itex]\sqrt{x^2+a^2+\sqrt{2}ax}[/itex] + [itex]\frac{a}{\sqrt{2}}[/itex] ln |[itex]\sqrt{a^2+x^2}[/itex] + x |

    Which is the right answer i think. thanks a lot for the help :)
     
  10. Nov 23, 2011 #9
    If i do the U substitution, I don't see how i can write x in terms of u
     
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