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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    $\begingroup$ I have no idea what the mark scheme is, but doing a simple fraction decomposition of your rational function always works! $\endgroup$ – Mariano Suárez-Álvarez Dec 23 '10 at 14:32
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    $\begingroup$ $\frac{v}{g-2v}=\frac{(g/2-v)-(g/2-v)+v}{g-2v}=\frac{g/2-v+v}{g-2v}-\frac12$ $\endgroup$ – J. M. isn't a mathematician Dec 23 '10 at 14:45
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    $\begingroup$ To find it, you have to think that you know how to integrate $\frac{dv}{a-bv}$, for $a$ and $b$ constants, but the $v$ in the numerator is a problem. So how to remove it? $\endgroup$ – Ross Millikan Dec 23 '10 at 15:34
  • $\begingroup$ @Ross Absolutely... not quite cracked that question yet though :P $\endgroup$ – 8128 Dec 23 '10 at 16:35
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Dec 23 '10 at 16:49

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

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  • $\begingroup$ 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. $\endgroup$ – 8128 Dec 24 '10 at 7:59

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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  • $\begingroup$ I'm afraid I don't understand what you are doing here at all. $\endgroup$ – 8128 Dec 23 '10 at 16:24
  • $\begingroup$ We want to make the $v$ on the LHS of the second line identically equal to the RHS. Thus the only choice for the constant $A$ is $-1/2.$ The $g-2v$ is on the RHS because it's in the denominator. After you have $A,$ $B$ follows since $B+Ag$ must equal $0.$ Does this help? $\endgroup$ – Derek Jennings Dec 23 '10 at 16:33
  • $\begingroup$ 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. $\endgroup$ – Derek Jennings Dec 23 '10 at 16:48

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