\(\displaystyle\int{\frac{{{f}}}{{{f}'}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{{\left({\ln{{f}}}\right)}'}}}{\left.{d}{x}\right.}=?\) I recently saw in another post that

rhedynogh0rp

rhedynogh0rp

Answered question

2022-03-28

ffdx=1(lnf)dx=?
I recently saw in another post that (lnf)=ff
where f=f(x)
From which it follows that ffdx=(lnf)dx=lnf+C
What about integrating the inverse of this?
I.e. what about ffdx=1(lnf)dx=?

Answer & Explanation

Theodore Davila

Theodore Davila

Beginner2022-03-29Added 14 answers

ffdx=(lnf)dx=lnf+C
is actually just a substitution: u=f(x);du=f(x)dx
About the second question:
ffdx=1(lnf)dx
Suppose that such formula exists. Let g(x) be any function and let
f(x)=eg(x)
Then the above formula would yield a general formula for
1g(x)dx.
Conversely if a formula for 1g(x)dx. exists then you can get your
formula by defining g(x)=ln|f(x)|.
The question you ask is equivalent to the existence of a formula for
1g(x)dx.
I highly doubt that this is true, but couldn't find the right transcendent function, I am sure someone smarter will ;)

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