If we start with $\int\frac{v}{g-2v} dv$ how would we go about integrating it? (g is a constant)
The answers to the past exam paper I have tell me to rearrange it to $\frac{1}{2}\int ( -1 + \frac{g}{g-2v} ) dv$ which integrates to $\frac{1}{2}v-\frac{1}{4}g\ln(g-2v) + c$
I can see that the rearrangement works 'in reverse', but I'm not sure how I would go about rearranging it the right way round.
Note: Part of question 3(iii) in January 2006 MEI Differential Equations paper.