# Help with the integral $\int \frac{1}{e - e^y} dy$

I'm having troubles trying to solve this one variable integral: $$\int \frac{1}{e - e^y} dy.$$

I've tried to apply the substitution method but it does not seems to work since I get something like: $$\int \frac{1}{u} \cdot \frac{du}{- e^y}.$$

I've put some tought on it, trying to find some algebric way to make it work, but without success. Can anyone show me how I can solve it? Thanks.

Hint:

$$\int \frac{1}{e - e^y} dy = \frac1e\int \frac{(e-e^y) + e^y}{(e - e^y)} dy$$

Hint: Take $u=e^{y}$. We get $$\int\frac{1}{e-e^{y}}dy=\int\frac{1}{\left(e-u\right)u}du=\frac{1}{e}\int\frac{1}{u}du-\frac{1}{e}\int\frac{1}{u-e}du.$$ Can you take from here?

• Sorry, i did not understood what did you do from step 2 to step 3 Commented Jun 13, 2016 at 20:11
• @FelipeChabatura I used the partial fractions decomposition. Commented Jun 13, 2016 at 20:55
• oh, i didn't learn how to use that yet, sorry Commented Jun 13, 2016 at 21:38

This equals $$\int\frac1u\frac{du}{u-e}$$ Do you know how to use partial fractions?

• No, i dont, sorry... Commented Jun 13, 2016 at 20:10

Note that we can write

\begin{align} \int \frac{1}{e-e^y}\,dy&=\int \frac{e^{-y}}{ee^{-y}-1}\,dy\tag 1\\\\ &=\bbox[5px,border:2px solid #C0A000]{-\frac{1}{e}\log|ee^{-y}-1|+C} \tag 2 \end{align}

In arriving at $(1)$, we simply multiplied the integrand by $\frac{e^{-y}}{e^{-y}}$.

In going from $(1)$ to $(2)$, we used the substitution $u=e^{-y}$ and noted the resulting integral was of the form $-\int \frac{1}{eu-1}\,du$

• Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark Commented Aug 12, 2016 at 1:55