# Integrating a differential

Gold Member

## Homework Statement

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.

## The Attempt at a Solution

Clueless.

tiny-tim
Homework Helper
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

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