# Integration of a fraction

1. Nov 7, 2011

### syj

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

i need to integrate:
$\frac{1}{r(u-r)}dr$

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. Nov 7, 2011

### Dick

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

3. Nov 7, 2011

### syj

ok
so i get
$\frac{1}{r(u-r)}=\frac{A}{r}+\frac{B}{u-r}$

so that
$B-A=0$
and
$Au=1$

so i got
$A=\frac{1}{u}=B$

this gives me
$\frac{1}{ur}+\frac{1}{u(u-r)}$

Last edited by a moderator: Nov 7, 2011
4. Nov 7, 2011

### HallsofIvy

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

5. Nov 8, 2011

### syj

$\int(\frac{1}{ur}+\frac{1}{u(u-r)})$

$=\frac{1}{u}\int\frac{1}{r}+\int\frac{1}{u-r}$

$=\frac{1}{u}(ln(r)-ln(u-r))$

$=\frac{1}{u}(ln\frac{r}{u-r})$

my problem is that i dont know where to put $r_0$ here. The solution has $r_0$

6. Nov 8, 2011

### HallsofIvy

Staff Emeritus
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. Nov 8, 2011

### syj

oh, sorry what i mean is that when it integrates they have $r_0$, sorry, i meant the solution has $r_0$