# Integrating a differential

1. Apr 1, 2012

### fluidistic

1. The problem statement, all variables and given/known data
There's a step I don't really understand in some "proof".
$d \left ( \frac{\mu }{T} \right )=-\frac{3R du}{2u}-\frac{Rdv}{v}$. Now he integrates both sides to get $\frac{\mu}{T}- \left ( \frac{\mu}{T} \right ) _0=-\frac{3R}{2} \ln \frac{u}{u_0}-R \ln \frac{v}{v_0}$.
I don't understand how an integration leads to such an expression.

2. Relevant equations

3. The attempt at a solution Clueless.

2. Apr 1, 2012

### tiny-tim

hi fluidistic!

this is standard …

if we integrate dx from xo to x1, we get x1 - xo

we then put x = x1, and it becomes ∫ dx = x - xo

they've done it for x = µ/T, dx = du/u, and dx = dv/v

3. Apr 1, 2012

### fluidistic

Thanks a lot. Crystal clear to me now. I can proceed further.