$\int\frac{f}{f'}\mathrm dx=\int\frac{1}{(\ln{f})'}\mathrm dx=\ $? I recently saw in another post that $(\ln{f})'=\frac{f'}{f}$
where $f=f(x)$
From which it follows that $\int\frac{f'}{f}\mathrm dx=\int(\ln{f})'\mathrm dx=\ln{f}+C$
What about integrating the inverse of this?
I.e. what about $\int\frac{f}{f'}\mathrm dx=\int\frac{1}{(\ln{f})'}\mathrm dx=?$
 A: $$\int\frac{f'}{f}dx=\int(\ln{f})'dx=\ln{f}+C$$ is actually just a substitution :$u=f(x) \,;\, du =f'(x) dx$.
About the second question: 
$$\int\frac{f}{f'}dx=\int\frac{1}{(\ln{f})'}dx$$
Suppose that such formula exists. Let $g(x)$ be any function and let $f(x)=e^{g(x)}$. 
Then the above formula would yield a general formula for 
$$\int \frac{1}{g'(x)} dx \,.$$
Conversely if a formula for $\int \frac{1}{g'(x)} dx \,.$ 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 
$$\int \frac{1}{g'(x)} dx \,.$$
I highly doubt that this is true, but couldn't find the right transcendent function, I am sure someone smarter will ;)
A: We give an explicit example that completes the argument of @user9176. Note first that $\dfrac{e^x}{x}$ does not have an elementary antiderivative. For a 
proof, see this. But in the notation of @user9176, 
$$\frac{e^x}{x}=\frac{1}{g'(x)},$$
where $g'(x)=xe^{-x}$.  Since  $\int xe^{-x}\,dx=-(xe^{-x}+e^{-x})+C$, the function $g(x)$ is an elementary function, and therefore so is $f(x)$, where $f(x)=e^{g(x)}$.
A: I wonder if you know about change of variables for an indefinite integral. You can read about it on the linked website, but the main idea is that
$$
\int G'(f(x))f'(x)dx = G(f(x))+c.
$$
In your case, $G(f) = \log f$ so $\displaystyle{G'(f) = \frac1f}$ and hence $\displaystyle{\frac{f'}f = G'(f)f'}$.
On the other hand, there is no such a function $G$ that $\displaystyle{G(f)f' = \frac{f}{f'}}$ for any function $f$ which is, say differentiable. The naive explanation is that if such function would exist then $\displaystyle{G(f) = \frac f{f'^2}}$ which is impossible since the LHS depends only on $f$ while RHS depends on both $f$ and $f'$.
