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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.

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    That was the point of Derek Jennings' response. You are basically dividing the numerator by the denominator as polynomials with$a$remainder term. Given P(v)/Q(v) with Q linear, you can write it as R(v)+b/Q(v) for$b$constant. And you know how to integrate both of these.2010-12-23

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Couldn't one just do a u-substitution with u=g-2v?

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    This was what I eventually decided on - was just going to write it up but you beat me to it. Whilst I'm sure the other answer is perfectly valid this fits far better with the techniques I am used to using.2010-12-24
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Here's how you can find your numerator. Set

$ v \equiv B + A(g-2v).$

This immediately gives $A = -1/2 $ and hence $ B = g/2 . $

Try $\int \frac{5g - 3v}{g+v} \textrm{ d}v$ to check that you have the idea.

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    Try a similar thing with the other example I gave: $5g-3v \equiv \ldots $ Ask yourself what you'd put on the RHS in this case, following the previous pattern.2010-12-23