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
$$\int\frac{f(x)}{f'(x)}dx=\int\frac{f}{\frac{df}{dx}}dx=\int f\frac{dx}{df}dx=\int f\frac{dx}{df}\frac{dx}{df}df$$ $$\int\frac{f(x)}{f'(x)}dx=\int \left(\frac{dx}{df}\right)^2f\:df$$ 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.}$$