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Integral dx/sqrt(x - a)integral dx/sqrt(1/ax)

  1. Jul 30, 2005 #1
    integral dx/sqrt(x - a)
    integral dx/sqrt(1/ax)
    Last edited: Jul 30, 2005
  2. jcsd
  3. Jul 30, 2005 #2
    reading suggestions
  4. Jul 30, 2005 #3
    sorry for wasting time I think I just understood substitution, out of embarassment im never coming back to this forum(bright side of life I learned more in three days than I have in two years)
  5. Jul 30, 2005 #4


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    We all embarrass ourselves at one time or another! (Some of us more than others. Believe me, I know about embarrassing myself!) PLEASE come back to PhysicsForums!

    (I need someone to make me look good!!!)

    Anyway, for those who are still wondering how to do these:

    [tex]\int \sqrt{x-a}dx[/tex]

    Let u= x-1 so du= dx and [tex]\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}[/tex]. The integral becomes [tex]\int u^{\frac{1}{2}}du[/tex] which can be done with the power law.

    [tex]\int \frac{dx}{\sqrt{\frac{1}{ax}}}[/tex]

    [tex]\frac{1}{\sqrt{\frac{1}{ax}}}[/tex] is just [tex]\sqrt{ax}[/tex].

    Let u= ax so du= adx, dx= (1/a)du and [tex]\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}[/tex]

    The integral becomes [tex]\frac{1}{a}\int u^{\frac{1}{2}}du[/tex]
    Last edited: Jul 31, 2005
  6. Jul 31, 2005 #5

    HallsofIvy, should not this equation be:
    [tex]\int \frac{dx}{\sqrt{x - a}} = \int u^{-\frac{1}{2}} du[/tex]
  7. Aug 1, 2005 #6


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    Oh, you're right! I missed the "1/" in the first post. Of course, the substitution would be exactly the same and the u-integral what you show. Thanks.
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