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Integration help

  1. Jul 30, 2008 #1
    how do i integrate (1+Z)/(2-2Z)
     
  2. jcsd
  3. Jul 30, 2008 #2
    What have you tried so far?
     
  4. Jul 30, 2008 #3

    HallsofIvy

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    [tex]\int\frac{1+Z}{2- 2Z}dZ= \frac{1}{2}\int\frac{1+Z}{1- Z}dZ[/tex]
    Does that help?

    and
    [tex]\frac{1+ Z}{1- Z}= -1- ?[/tex]
     
  5. Jul 30, 2008 #4

    Dick

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    Try a u substitution. What's the obvious one?
     
  6. Jul 30, 2008 #5
    substitute 2-2Z for u? but then ive still got something over something else and still dont get what method to use :S
     
  7. Jul 30, 2008 #6

    Defennder

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    As HallsOfIvy pointed out, you can just use polynomial long division to decompose the fraction into something more integrable.
     
  8. Jul 30, 2008 #7

    Dick

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    If u=2-2Z then Z=(2-u)/2. Substitute that into 1+Z. Now it's all u's.
     
  9. Jul 30, 2008 #8
    Dick said it all. You need to write Z in terms of u, then you need to find dz/du by differentiating that. You substitute everything in and integrate.
     
  10. Jul 31, 2008 #9

    HallsofIvy

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    But you will still have to divide the fraction. Seems simpler just to reduce the original fraction.
     
  11. Jul 31, 2008 #10
    If you factor out the 2, sub u = 1-z and z = 1-u, all the work has been done for you. Just seperate the fraction and integrate.
     
  12. Jul 31, 2008 #11

    HallsofIvy

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    And, one more time, why not just "separate the fraction" (by which I assume you mean "divide") without the substitution?
     
  13. Jul 31, 2008 #12

    Air

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    If you substitute [itex]u=2-2z[/itex], and you simplify it, you will end up with this at one point:

    [itex]-\frac12 \int \left( \frac{2}{u} - \frac12 \right) \ \mathrm{d}u \implies \frac14 \int 1 \ \mathrm{d}u - \int \frac{1}{u} \ \mathrm{d}u[/itex]

    Integrate this and remember to substitute [itex]u=2-2z[/itex] in the end to get the final answer.
     
  14. Aug 1, 2008 #13

    pps

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    [tex] u = z-1 [/tex]
    [tex] du = dz [/tex]
    [tex] z = u + 1 [/tex]

    [tex]\frac{1}{2} \int \frac{1+z}{1-z}dz =\frac{1}{-2} \int \frac{1+z}{z-1}dz = [/tex]

    [tex]\frac{1}{-2} \left( \int \frac{dz}{z-1} + \int \frac{z}{z-1}dz \right) = [/tex]

    [tex]\frac{1}{-2} \left( \int \frac{du}{u} + \int \frac{u + 1}{u} du \right) = [/tex]

    [tex]\frac{1}{-2} \left( \int \frac{du}{u} + \int \frac{du}{u} + \int du \right) = [/tex]

    Continue from this point.
     
  15. Aug 1, 2008 #14

    HallsofIvy

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    And if you use the fact that [tex]\frac{1+z}{2-2z}= -\frac{1}{2}\left(1+ \frac{1}{1-z}\right)[/tex] you have
    [tex]\int \frac{1+z}{2- 2z}dz= -\frac{1}{2}\int (1+ \frac{1}{z-1}dz= -\frac{1}{2}(z+ ln|z-1|)+ C[/tex]

    That seems simpler to me.
     
  16. Aug 1, 2008 #15

    pps

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    Your method is indeed much simpler. =)
     
  17. Aug 2, 2008 #16

    Dick

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    I don't think it's MUCH simpler. It's six of one and half a dozen of the other. Neither one is all that complicated. If you ask me. And you don't have to.
     
  18. Aug 2, 2008 #17
    That's false though... Just check z=0 the left side gives 1/2, but the right side gives -(1/2)(2)=-1...
     
  19. Aug 2, 2008 #18

    Dick

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    At this point I'm wondering who cares. franky2727 hasn't checked in since the initial post. I guess this thread is just getting a lot of attention because nothing else is going on.
     
  20. Aug 2, 2008 #19

    HallsofIvy

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    Typo: I meant
    [tex]-\frac{1}{2}\left(1+ \frac{2}{z-1}\right)[/tex]
     
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