How do you integrate with respect to a function?

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SUMMARY

The discussion centers on solving the integral equation \(\int \frac{1}{f'(x)} df(x) = g(x)\) for \(f(x)\). The solution provided by Wolfram-Alpha, \(f(x) = c_1 e^{\int \frac{1}{g(\xi)} d\xi}\), is questioned for its validity, particularly regarding its applicability to all functions \(g\). The user highlights that the solution seems overly restrictive and does not account for all differentiable functions \(f\) with a non-zero first derivative. The user seeks clarification on the integration with respect to a function and its definition.

PREREQUISITES
  • Understanding of integral calculus and differential equations
  • Familiarity with the concept of differentials, specifically \(d(f(x))\)
  • Knowledge of the properties of differentiable functions
  • Experience with mathematical software like Wolfram-Alpha
NEXT STEPS
  • Study the derivation of the integral equation \(\int \frac{1}{f'(x)} df(x) = g(x)\)
  • Learn about the properties of differentiable functions and their integrals
  • Explore the concept of integration with respect to a function in more depth
  • Review the limitations of solutions provided by computational tools like Wolfram-Alpha
USEFUL FOR

Students and educators in calculus, mathematicians exploring differential equations, and anyone interested in the nuances of integration techniques.

Char. Limit
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Homework Statement



So I thought of an interesting problem, and here it is:

Solve \int \frac{1}{f'(x)} df(x) = g(x) for f(x).

Now, I checked Wolfram-Alpha to see if an answer existed, and they gave me this:

f(x) = c_1 e^{\int \frac{1}{g(\xi)} d\xi}

But, you know that Wolfram-Alpha doesn't show steps. So I want to know how they got from start to finish. I checked the solution, and it seems to work, but how did they get there?

Note that I have taken all three Calculus classes, but no differential equations experience.

EDIT: As a further question, how do you integrate with respect to a function? Is that even defined?
 
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d(f(x)) is the differential of f(x). It's f'(x)*dx. That makes the left side pretty easy to integrate.
 
Dick said:
d(f(x)) is the differential of f(x). It's f'(x)*dx. That makes the left side pretty easy to integrate.

It also means that the solution contains any function f that is differentiable and has a non-zero first derivative, and any function g, as long as that function is of the form g(x)=c.

Wolfram's solution does look a bit weird though.
If you substitute g(x)=c, it works out to f(x)=c1.exp(ln(x))=c1.x.
It seems to me that Wolfram's answer is wrong.
It allows for any g, which is not true.
And it finds a specific solution for f, which is too restrictive.

Funny though =).
 

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