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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.

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  • $\begingroup$ Which substitutions have you tried? $\endgroup$ – Travis Willse Apr 19 '17 at 13:40
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$$\begin{array}{rcl} \displaystyle \int e^{-x} \frac{f'(e^{-x})}{f(e^{-x})} \ \mathrm dx &=& \displaystyle -\int \frac{f'(e^{-x})}{f(e^{-x})} \ \mathrm de^{-x} \\ &=& \displaystyle -\int \frac{f'(u)}{f(u)} \ \mathrm du \\ &=& \displaystyle -\int \frac{1}{f(u)} \ \mathrm df(u) \\ &=& \displaystyle -\ln f(u) + C \\ &=& \displaystyle -\ln f(e^{-x}) + C \\ \end{array}$$

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

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

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