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Help with integration

  1. Oct 2, 2008 #1
    Alright so I have a math problem that I have broken down into parts and one of my parts I am having trouble with is as follows.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I attempted to solve it by making z=x2-1 and dz=2xdx. From here I am just no sure how to go about breaking this part up into something I can use. Any steers into the right direction? Thanks guys, your help really is greatly appreciated.
  2. jcsd
  3. Oct 3, 2008 #2
    I assume you didn't mean to use a contour integral. Is this the equation you meant?

    [tex]\int{\frac{x}{\sqrt{x^{2}-1}}dx}[/tex] It should be a simple substitution problem as you were doing.
  4. Oct 3, 2008 #3
    yes that is what I meant except outside of the [tex]\sqrt{}[/tex] there is another x , that substitution is part of a much larger master problem but I have the rest figured out, I just need a helpful push into the right direction as far as solving this piece, thank you :).
  5. Oct 3, 2008 #4
    I took one x out of the numerator and one out of the denominator. Are you sure what I posted isn't what you meant because that's all I did to simplify it algebraically.
    Last edited: Oct 3, 2008
  6. Oct 3, 2008 #5
    tuche haha still any help ?
  7. Oct 3, 2008 #6
    Well you have a simple integral after you do your z substitution. [tex]\frac{1}{2}\int{\frac{dz}{\sqrt{z}}dz}=?[/tex]
  8. Oct 3, 2008 #7

    Gib Z

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    I think you meant

    [tex]\frac{1}{2} \int \frac{ \frac{dz}{dx}}{\sqrt{z}} dx[/tex].
  9. Oct 3, 2008 #8
    so after putting the substitution back in I woild have 1/2[tex]\int[/tex]2x/[tex]\sqrt{x2-1}[/tex] correct?
  10. Oct 3, 2008 #9


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    Staff Emeritus
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    That is exactly the same as the
    [tex]\frac{1}{2} \int\frac{dz}{\sqrt{z}}[/tex]

    That is not what you originally posted! You originally posted
    [tex]\int \frac{x^2 dx}{x\sqrt{x^2- 1}}[/tex]
    Which is simply the same as
    [tex]\int \frac{x dx}{\sqrt{x^2- 1}}[/tex]
    because you can cancel an "x" in numerator and denominator. Now, let z= x2- 1.
  11. Oct 3, 2008 #10

    Gib Z

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    Homework Helper

    Yes it is, but different to:

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