Question

(x^2+y^2+2x)dx+2xydy=0​

Answer 1

$\displaystyle(x^2+y^2+2x)dx+2xydy=0\\\begin{cases}\displaystyle\frac{\delta F}{\delta x}=x^2+y^2+2x\\\displaystyle\frac{\delta F}{\delta y}=2xy\end{cases}\\F=\int 2xydy=xy^2+\phi(x)\\\frac{\delta F}{\delta x}=y^2+\phi'(x)=x^2+y^2+2x\\\phi'(x)=x^2+2x\\\phi(x)=\int(x^2+2x)dx=\frac{x^3}{3}+x^2+C\\F=\frac{x^3}{3}+x^2+xy^2+C$