- #1

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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.

- Thread starter syj
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- #1

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i need to integrate:

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

u is a constant

im not sure if i should decompose the fraction. i tried that, but it didnt seem to be of any help.

- #2

Dick

Science Advisor

Homework Helper

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Yes, use partial fractions. Show us how it didn't help. It should have.

- #3

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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]

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:

- #4

HallsofIvy

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Good! Now, what do you get when you intgrate those?

- #5

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[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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