First Order ODE/x dy = (x^5 + x^3 y^2 + y) dx

Theorem
The first order ODE:
 * $(1): \quad x \rd y = \paren {x^5 + x^3 y^2 + y} \rd x$

has the general solution:
 * $\arctan \dfrac x y = -\dfrac {x^4} 4 + C$

Proof
Rearranging, we have:
 * $y \rd x - x \rd y = -\paren {x^2 + y^2} x^3 \rd x$

from which:
 * $\dfrac {y \rd x - x \rd y} {x^2 + y^2} = - x^3 \rd x$

From Differential of Arctangent of Quotient:
 * $\map \d {\arctan \dfrac x y} = \dfrac {y \rd x - x \rd y} {x^2 + y^2}$

from which $(1)$ evolves into:
 * $\map \d {\arctan \dfrac x y} = -x^3 \rd x$

Hence the result:
 * $\arctan \dfrac x y = -\dfrac {x^4} 4 + C$