Integral with radical denominator.

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Homework Statement




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.
 

Answers and Replies

  • #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?
 
  • #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.
 
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  • #4
Dick
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I would try writing [itex]x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a)
=(x-\sqrt a)^2)+(a^2-a)[/itex] and see if that leads anywhere.

[Edit] I see Dick types faster than I do...

No, we type about the same. I just say less.
 
  • #5
SammyS
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I would try writing [itex]x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a)
=(x-\sqrt a)^2)+(a^2-a)[/itex] and see if that leads anywhere.

[Edit] I see Dick types faster than I do...

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]
 
  • #6
LCKurtz
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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]

Yes, thanks. I edited it (about 3 times :frown:) to correct it.
 
  • #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.
 
  • #8
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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 :)
 
  • #9
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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.

If i do the U substitution, I don't see how i can write x in terms of u
 

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