The solution of (x+y+1)dy=dx are:

Put x + y + 1 = t; $1+\frac{dy}{dx}=\frac{dt}{dx}1⇒\frac{dt}{dx}=(\frac{1}{t}+1)⇒\int\frac{-1+(1+t)dt}{1+t}=\int dx$

$⇒-ln(1+t)+t=x+c'$

$⇒-ln(x+y+2)+y+1=c'⇒x+y+2=ce^y$

Or log (x + y + 2)= c₁ + y where c₁ = ln c