Answer
$$y=e^{-x } \ln |e^x-1|+ce^{-x } $$
Work Step by Step
Given $$ \frac{dy}{dx} + y+\frac{1}{1-e^x}=0$$
Rewriting equation
$$ \frac{dy}{dx} + y=\frac{1}{e^x-1}$$
Since
\begin{align*}
\mu(x)&=e^{\int dx}\\
&=e^{x }
\end{align*}
Then
\begin{align*}
y\mu(x)&=\int \mu(x) q(x)dx\\
e^{x } y&=\int \frac{e^{x }}{e^x-1} dx\\
&=\ln |e^x-1|+c
\end{align*}
Hence $$y=e^{-x } \ln |e^x-1|+ce^{-x } $$
