y'=cos(x+y+1)
Let z=x+y+1 and z'=dz/dx=1+y'=1+cos(x+y+1)=1+cosz.
dz/dx=1+cosz; ∫dz/(1+cosz)=∫dx=x.
x=∫dz/(1+cosz); cosz=2cos^2(z/2)-1, so x=½∫sec^2(z/2)dz=½(2tan(z/2))=tan(z/2)+C.
Therefore x=tan((x+y+1)/2)+C (implicit equation).
CHECK
Differential of the solution: 1=½sec^2((x+y+1)/2)(1+y')=(1+y')/(2cos^2((x+y+1)/2)=
(1+y')/(1+cos(x+y+1)), and 1+y'=1+cos(x+y+1), so dy/dx=cos(x+y+1).
The check demonstrates that x=tan((x+y+1)/2)+C is a solution in implicit form.
So tan((x+y+1)/2)=x-C and x+y+1=2arctan(x-C), y=2arctan(x-C)-x-1 (explicit form).