Solution of the differential equation $\frac{dy}{dx}=1+x+y+x y$ are : Where C is an arbitrary constant

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

$\log (1+y)=x+\frac{x^2}{2}+C$

The correct answer is Option (3) → $\log (1+y)=x+\frac{x^2}{2}+C$