Question: Solve $(e^x+1)\,y\,dy=(y+1)\,e^x\,dx$.

Solution:

Separate: $\dfrac{y}{y+1}\,dy=\dfrac{e^x}{e^x+1}\,dx$

Integrate: $y-\ln(y+1)=\ln(e^x+1)+C$

Answer (implicit): $\boxed{\,y-\ln(1+y)=\ln(1+e^x)+C\,}$

Equivalently: $\dfrac{e^{y}}{y+1}=K(1+e^x)$.

Answer : $e^y=k(y+1)(1+e^x)$