Integral of function over its derivative I am looking for a solution to the following integral:
For a function $f(x)$,
$$ \int \frac{f}{f'}dx, $$
where $f'$ is the derivative of $f$ with respect to $x$.
It is clear that $\int \frac{f'}{f} dx = \log(f)$, but I have no idea how to solve the above one.
Any help would be greatly appreciated!!
Cheers,
Marc
 A: $$\int\frac{f(x)}{f'(x)}dx=\int\frac{f}{\frac{df}{dx}}dx=\int f\frac{dx}{df}dx=\int f\frac{dx}{df}\frac{dx}{df}df$$
$$\int\frac{f(x)}{f'(x)}dx=\int \left(\frac{dx}{df}\right)^2f\:df$$
Consider the inverse function of $f(x)$ say $x=g(f)$
$$\boxed{\int\frac{f(x)}{f'(x)}dx=\int \left(\frac{dg(f)}{df}\right)^2f\:df}$$
This is a function of the variable $f$ that has to be integrated in the common sens.
$$$$
EXAMPLE :
$$f(x)=e^{2x}+1$$
The inverse function is $$x=\frac12\ln|f-1|=g(f)$$
$$\frac{dg}{df}=\frac{1}{2(f-1)}$$
$$\int\left(\frac{dg(f)}{df}\right)^2f\:df=\int\frac{1}{4(f-1)^2}f\:df=-\frac{1}{4(f-1)}+\frac14\ln|f-1|$$
$f-1=e^{2x}$
$$\int\left(\frac{dg(f)}{df}\right)^2f\:df=-\frac14 e^{-2x}+\frac12 x$$
To be compared to :
$$\int\frac{f(x)}{f'(x)}dx=\int \frac{e^{2x}+1}{2e^{2x}}=\int\left(\frac12+2e^{-2x} \right)dx=\frac12 x-\frac14e^{-2x}$$
$$\text{Thus the equality}\quad\int\frac{f(x)}{f'(x)}dx=\int \left(\frac{dg(f)}{df}\right)^2f\:df=\frac12 x-\frac14e^{-2x}\quad\text{is satisfied.}$$ 
