Solve (y^2+z^2+2xy+2xz) dx; +(x^2+z^2+2xy+2yz) dy…

Solve \begin{align*} &(y^2+z^2+2xy+2xz)\,dx\\ &+(x^2+z^2+2xy+2yz)\,dy\\ &+(x^2+y^2+2xz+2yz)\,dz = 0. \end{align*}

Step 1 · Check exactness

$P_y=2y+2x=Q_x$, $P_z=2z+2x=R_x$, $Q_z=2z+2y=R_y$. \checkmark Exact.

Step 2 · Find the potential $F$

Integrating $F_x=P$ gives $F=xy^2+xz^2+x^2y+x^2z+g(y,z)$; matching $F_y=Q$ gives $g_y=z^2+2yz$, so $g=yz^2+y^2z+h(z)$; matching $F_z=R$ gives $h'(z)=0$. Hence \[ F = x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = x^2(y+z)+y^2(x+z)+z^2(x+y). \]

Step 3 · Recognise the factored form

Expanding $(x+y)(y+z)(z+x)$ gives $F+2xyz$, so equivalently \[ (x+y)(y+z)(z+x)-2xyz = C. \]

Verification

$F_x=y^2+z^2+2xy+2xz=P$, $F_y=x^2+z^2+2xy+2yz=Q$, $F_z=x^2+y^2+2xz+2yz=R$. \checkmark