Integrating a Fraction with a Constant: Decomposition or Not?

[syj]

Homework Statement

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

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.

 

[Dick]

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

 

[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)}

 

[HallsofIvy]

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

 

[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 don't know where to put r_0 here. The solution has r_0

 

[HallsofIvy]

Then perhaps you should have told us from the start what the problem really is. The problem you posted does not have any r₀.

 

[syj]

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