The solution of the differential equation $(x+1) \frac{d y}{d x}=1+y$ is

$\frac{1+y+y^2}{1+x^2}=C$ $\log (x+1)-\log \left(y+\frac{1}{2}\right)=C$ $\frac{x+1}{y+1}=C$ $\log (1+y)-\frac{\sqrt{3}}{2} \log (x+1)=C$

$\frac{x+1}{y+1}=C$

$(x+1) \frac{d y}{d x}=1+y \Rightarrow \int \frac{1}{1+y} d y=\int \frac{1}{1+x} d x$ $\Rightarrow \log |1+y|=\log |1+x|+\log |c|$ taking logarithmic constant for convenience so  $\log |1+y|=\log |c(1+x)|$ $\Rightarrow \frac{1+y}{1+x}=c$  or  $\frac{1+x}{1+y}=c$ Option: 3