Is this integral solvable with some substitution?

Question

Is there a general way to solve analytically the following type of integral?

$$\int e^{-x} \frac{f'(e^{-x})}{f(e^{-x})}dx$$

I am thinking about some substitution but I don't see a direct result.

Answer 1

$$\begin{array}{rcl} \displaystyle \int e^{-x} \frac{f'(e^{-x})}{f(e^{-x})} \ \mathrm dx &=& \displaystyle -\int \frac{f'(e^{-x})}{f(e^{-x})} \ \mathrm de^{-x} \\ &=& \displaystyle -\int \frac{f'(u)}{f(u)} \ \mathrm du \\ &=& \displaystyle -\int \frac{1}{f(u)} \ \mathrm df(u) \\ &=& \displaystyle -\ln f(u) + C \\ &=& \displaystyle -\ln f(e^{-x}) + C \\ \end{array}$$

Answer 2

Hint:$$\int e^{-x} \frac{f'(e^{-x})}{f(e^{-x})}dx=-\int \frac{(-e^{-x})f'(e^{-x})}{f(e^{-x})}dx$$

Answer 3

Hint: Note that by the chain rule$$\frac d{dx} \log f(e^{-x}) = \frac{f'(e^{-x})}{f(e^{-x})}(-e^{-x})$$