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

Theorem
The first order ODE:
 * $(1): \quad x \, \mathrm d y = \left({x^5 + x^3 y^2 + y}\right) \mathrm d x$

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

Proof
Rearranging, we have:
 * $y \, \mathrm d x - x \, \mathrm d y = -\left({x^2 + y^2}\right) x^3 \, \mathrm d x$

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

From Differential of Arctangent of Quotient:
 * $\mathrm d \left({\arctan \dfrac x y}\right) = \dfrac {y \, \mathrm d x - x \, \mathrm d y} {x^2 + y^2}$

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

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