The general solution of the differential equation $(e^x+1)ydy=(y+1)e^xdx$ is

Question

$$(A)\;(y+1)=k(e^x+1) \quad (B)\;y+1=e^x+1+k \quad (C)\;y=log\{k(y+1)(e^x+1)\} \quad (D)\;ylog{\frac{e^x+1}{y+1}}+k$$