Integrating a Fraction with a Constant: Decomposition or Not?

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Homework Help Overview

The discussion revolves around the integration of the function \(\frac{1}{r(u-r)}\) with respect to \(r\), where \(u\) is a constant. Participants are exploring whether to use partial fraction decomposition as part of their approach.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses uncertainty about the usefulness of decomposing the fraction. Some participants suggest using partial fractions and inquire about the original poster's attempts and results.

Discussion Status

Participants are actively discussing the integration process, with some providing guidance on the use of partial fractions. The original poster has shared their steps and is seeking clarification on the inclusion of \(r_0\) in the solution, indicating a productive exchange of ideas.

Contextual Notes

There is a mention of \(r_0\) in the context of the solution, which was not included in the initial problem statement. This raises questions about the completeness of the problem setup and the assumptions being made.

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.
 
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Yes, use partial fractions. Show us how it didn't help. It should have.
 
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:
Good! Now, what do you get when you intgrate those?
 
[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 don't know where to put [itex]r_0[/itex] here. The solution has [itex]r_0[/itex]
 
Then perhaps you should have told us from the start what the problem really is. The problem you posted does not have any r0.
 
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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