Answer
$\ln (1+e^y)=e^x+x+c$
Work Step by Step
Rewrite the given equation and then integrate.
We have $\dfrac{dy}{dx}=(e^{x}+1)(e^{-y}+1)$
This implies that $ \int (e^{x}+1) dx=\int \dfrac{1}{e^{-y}+1} dy$
Re-write as: $\int (e^x+1) dx=\int \dfrac{e^y}{1+e^{y}} dy $
Plug $a=e^y \implies da=e^y dy$
Now, we have $ \int (e^{x}+1) dx=\int \dfrac{da}{1+a}$
and $e^x+x+c=\ln (1+a)$
Thus, $\ln (1+e^y)=e^x+x+c$
