# Integration of a fraction

## Homework Statement

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

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.

## The Attempt at a Solution

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

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:
HallsofIvy
Homework Helper
Good! Now, what do you get when you intgrate those?

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

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

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

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

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

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