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I don't quite understand the steps involved in this indefinite integral with respect to $t$:

$$\int\frac1{M}\frac{dM}{dt}$$

The explanation I've does this substitution:

$$u = M(t)$$

Which seems to make the integral become:

$$\int\frac1{u}du$$

From here I understand that we integrate with the exponential rule, to get:

$$\ln|u| + C$$

My explanation then substitutes back $M$ to get:

$$\ln|M| + C$$

These are my understanding problems:

  1. What is the meaning of integrating $dM/dt$? I'm familiar with integrating functions with respect to one variable, like this: $$\int f(x)dx$$ but $dM/dt$ is the notation for a derivative? I just don't understand what we're doing here.

  2. I don't quite get why the substitution $$u = M(t)$$ works here. Probably relates to understanding problem 1.

  3. Why when we substitute $M(t)$ back we get: $$\ln|M| + C$$ instead of: $$\ln|M(t)| + C$$ the later looks strange, but we did $u = M(t)$, so my primitive understanding tells me to substitute back literally $M(t)$. Is the $(t)$ implicit?

Sebastiano
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User
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  • To whoever closed the question because it needs to be "more focused", I don't get it - this is about exactly one problem (calculating a specific indefinite integral). Am I supposed to open a new question per step where I need help? The understanding problems are also linked. This is nonsense. – User Feb 07 '23 at 07:04
  • Alternatively, sidestep the informal notation and rigorously derive indefinite integration as a theorem in just $3$ lines here. – ryang Feb 07 '23 at 16:40

1 Answers1

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Your confusion is understandable. The notation really should be written $$\int \frac{1}{M(t)} \frac{dM}{dt} \, \color{red}{dt}. \tag{1}$$ The reason becomes apparent if we pick a specific example; e.g., $$M(t) = \sin t.$$ Then the integral $(1)$ takes on the form

$$\int \frac{1}{\sin t} \cos t \, \color{red}{dt}. \tag{2}$$ Without the $dt$ (in red), the original notation would read $$\int \frac{1}{M} \frac{dM}{dt} = \int \frac{1}{\sin t} \cos t$$ and this is not really complete.

As for how the substitution works, you can also see it through the example I chose: we let $$u = \sin t, \quad du = \cos t \, dt, \tag{3}$$ which gives $$\int \frac{1}{\sin t} \cos t \, dt = \int \frac{1}{u} \, du = \log |u| + C = \log |\sin t| + C. \tag{4}$$

The same mechanism is valid for a generic differentiable function $M$:

$$u = M(t), \quad du = M'(t) \, dt = \frac{dM}{dt} \, dt. \tag{5}$$

The confusion over whether to use $M(t)$ or $M$ is more of a matter of notational brevity: when $M$ is used, it is implied through the term $dM/dt$ that $M$ is some function of $t$. Either $M$ or $M(t)$ is acceptable.

heropup
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