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Integration of a fraction

  1. Nov 7, 2011 #1

    syj

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

    i need to integrate:
    [itex]\frac{1}{r(u-r)}dr[/itex]

    2. Relevant equations

    u is a constant

    3. The attempt at a solution

    im not sure if i should decompose the fraction. i tried that, but it didnt seem to be of any help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 7, 2011 #2

    Dick

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    Homework Helper

    Yes, use partial fractions. Show us how it didn't help. It should have.
     
  4. Nov 7, 2011 #3

    syj

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    ok
    so i get
    [itex]\frac{1}{r(u-r)}=\frac{A}{r}+\frac{B}{u-r}[/itex]

    so that
    [itex]B-A=0[/itex]
    and
    [itex]Au=1[/itex]

    so i got
    [itex]A=\frac{1}{u}=B[/itex]

    this gives me
    [itex]\frac{1}{ur}+\frac{1}{u(u-r)}[/itex]
     
    Last edited by a moderator: Nov 7, 2011
  5. Nov 7, 2011 #4

    HallsofIvy

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    Good! Now, what do you get when you intgrate those?
     
  6. Nov 8, 2011 #5

    syj

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    [itex]\int(\frac{1}{ur}+\frac{1}{u(u-r)})[/itex]

    [itex]=\frac{1}{u}\int\frac{1}{r}+\int\frac{1}{u-r}[/itex]

    [itex]=\frac{1}{u}(ln(r)-ln(u-r))[/itex]

    [itex]=\frac{1}{u}(ln\frac{r}{u-r})[/itex]

    my problem is that i dont know where to put [itex]r_0[/itex] here. The solution has [itex]r_0[/itex]
     
  7. Nov 8, 2011 #6

    HallsofIvy

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    Then perhaps you should have told us from the start what the problem really is. The problem you posted does not have any r0.
     
  8. Nov 8, 2011 #7

    syj

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    oh, sorry what i mean is that when it integrates they have [itex]r_0[/itex], sorry, i meant the solution has [itex]r_0[/itex]
     
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