First Order ODE/y' = (x + y)^2

Theorem
The first order ODE:
 * $\dfrac {\d y} {\d x} = \paren {x + y}^2$

has the general solution:
 * $x + y = \map \tan {x + C}$

Proof
Make the substitution:
 * $z = x + y$

Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = b = 1$: