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

Theorem
The first order ODE:
 * $\dfrac {\mathrm d y} {\mathrm d x} = \left({x + y}\right)^2$

has the solution:
 * $x + y = \tan \left({x + C}\right)$

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

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