Integration of a fraction

  • Thread starter syj
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  • #1
syj
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Homework Statement



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

Homework Equations



u is a constant

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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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Yes, use partial fractions. Show us how it didn't help. It should have.
 
  • #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]
 
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  • #4
HallsofIvy
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Good! Now, what do you get when you intgrate those?
 
  • #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]
 
  • #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.
 
  • #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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