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Disclaimer: I kind of work through and understand the solution a bit as I typed this problem out and my question morphs from one of "how does the author get the solution they've presented?" to one of "Is this procedure standard for this type of problem?"

I'm working through a variable mass projectile example in a classical mechanics text involving the following integral $$\int^{v(z)}_0 \frac{dv}{\frac{g}{v} - \alpha v}$$ The author doesn't provide a step-by-step solution to the problem but precedes the solution to the integral by saying "Integrating (test by substitution) the velocity solution is" $$-\frac{1}{2\alpha}\ln(g - \alpha v^{2}) \vert^{v(z)}_0$$ I am a little confused as to how we got the solution and was hoping for some insight.

To me, this form of solution looks like a u-substitution where we got the integral in the form $\frac{du}{u}$ however when performing a u-substitution with $u = \frac{g}{v}- \alpha v$ we find that $du = -\frac{g}{v^2} - \alpha\ dv \implies dv = \frac{du}{-\frac{g}{v^2} - \alpha}$ so that our integral becomes $$\int \frac{du}{-\frac{u}{v}u}$$

I see that the argument of the natural logarithm is $v(\frac{g}{v} - \alpha v)$ so I suspect that the author carried out this integration with this u-substitution instead of using the original since this seems to work when considering $du = 0 - 2\alpha v dv$. Then the extra $v$ term gets canceled out by the fact that when transferring from the $v$ integral to the $u$ integral we have $\frac{u}{v}$ in the denominator.

Question: So, my question is, when would one use this method? Is it commonplace to use in integrals which involve polynomials where the lowest order term is $-1$? Would you use a similar method if the lowest order term was say $-3$ but instead, multiplying the original expression by $v^3$? Is this even a "method" or just a trick that happened to work for this one integral? This problem had me stumped for a while but typing this up I finally see the solution I just don't understand if this is some class of procedure for solving certain integrals that I've not been exposed to or where it came from, exactly.

Numerical Disintegration
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    My naive take: no one likes fractions within fractions, so we multiply the numerator and denominator by $v$, which yields $\int^{v(z)}_0 \frac{v , dv}{g - \alpha v^2}$. Then the numerator is the derivative of the denominator, so we do a $u$-substitution, and voilà, out pops the answer. There is a more general algorithm for integrating rational functions; see this post. – Viktor Vaughn Dec 03 '22 at 04:06

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I agree with your reasoning in the $u$-substitution part. The substitution $u=g-av^2$ can solve this integral. To answer your question, I would like to add some intermediate steps and show that it is nothing more than a usual $u$-substitution.

$\int\dfrac{dv}{\frac{g}{v}-\alpha v}$

$=\int\dfrac{vdv}{g-\alpha v^2}$

$=\int\dfrac{\frac{1}{2}dv^2}{g-\alpha v^2}$

$=\int\dfrac{-\frac{1}{2\alpha}d(g-\alpha v^2)}{g-\alpha v^2}=-\dfrac{1}{2\alpha}\ln|g-\alpha v^2|+C$

Note:

  1. In row 2, we can transform a negative-degree expression to a positive one by simplifying the fraction.

  2. In rows 3 & 4, the $u$-substitution is used twice. The reason why setting $u=g-\alpha v^2$ works is that we make $dv^2$ and $(g-\alpha v^2)$ have the same degrees, as in $\int\dfrac{1}{k-u}du$ for any real constant $k$.

So does this method work in general for all expressions with a negative-degree term in the denominator? Look at this:

Let $m,n>0$.

$\operatorname{\Large\int}\dfrac{1}{\dfrac{g}{v^m}-\alpha v^n}dv$

$=\operatorname{\Large\int}\dfrac{v^m}{g-\alpha v^{m+n}}dv$

$=\operatorname{\Large\int}\dfrac{\frac{1}{m+1}}{g-\alpha v^{m+n}}dv^{m+1}$

Here, we need $m+n = m+1$, i.e. $n=1$, in order to repeat the steps in the previous illustration.

Therefore, this method works for expressions in this form $\operatorname{\Large\int}\dfrac{1}{\dfrac{g}{v^m}-\alpha v}dv$, for any positive $m$.

Hirofumi Ryo
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