Consider a rocket in free space of initial mass (including fuel) of $m_0$ using up/ejecting fuel at a constant rate $r$ (mass per time). That means that at time $t$ the rocket has mass $m=m_0-rt$. The velocity of the exhaust is $v_e$.
When firing the rocket imparts $\dot{m}v_e$ impulse per unit time (force) on the exhaust (at first I fought this is where the problem lies but after more considerations it seams legit). This must be equal (and opposite) to the impulse imparted on the rocket during that time and hence the rate of change of momentum.
$$ -\dot{m}v_e = \frac{d}{dt} (mv) = \dot{m}v + m\dot{v}$$ substituting in $m=m_0-rt$ and $\dot{m}=-r$ we get $$ rv_e = -rv + (m_0-rt)\dot{v}$$ applying the reverse chain rule (I know I could have simply not expanded the term for momentum and gotten the same result) and integrating $$ rv_e = \frac{d}{dt} (m_0-rt)v$$ $$ trv_e +c= (m_0-rt)v$$ rearranging, determining the constant (for say $v(t=0)=0$) and substituting $t=t_{final}$ we get that: $$\frac{\Delta v}{v_e}=\frac{tr}{m_0-rt} = \frac{\text{fuel mass}}{\text{vehicle mass}}$$
which is quite different from the actually equation of $$\frac{\Delta v}{v_e} = \ln\left(\frac{m_0}{m_f}\right)$$
I understand the derivation of this equation, what I'm missing is at what point in the derivation did my method go wrong. I have a feeling that stating that the fuel is ejected at a constant mass rate trivialised some calculus and made me miss a term of factor which would give the exponential/logarithm, but thats just a hunch.
