Answer
$\ln (1+e^y)=e^x+x+c$
Work Step by Step
Given: $\dfrac{dy}{dx}=e^{x-y}+e^{x}+e^{-y}+1$
Re-arrange the given equation and integrate as follows:.
Then, $\dfrac{dy}{dx}=(e^{x}+1)(e^{-y}+1)$
Then $ \int (e^{x}+1) dx=\int \dfrac{1}{e^{-y}+1} dy \implies \int (e^x+1) dx=\int \dfrac{e^y}{1+e^{y}} dy $ ....(1)
Let us take the help of $u$ substitution.
plug $u=e^y \implies du=e^y dy$
Equation (1) becomes: $ \int (e^{x}+1) dx=\int \dfrac{du}{1+u}$
$e^x+x+c=\ln (1+u)$
Hence, $\ln (1+e^y)=e^x+x+c$
