Find a function that makes a differential form exact We have $A=\mathbb{R^3}\backslash \left\{ (0,0,z):z\in \mathbb{R}\right\}$ and $\omega$ the differential form:
$$\omega :=\left(\frac{4x^2+2zx}{x^2+y^2}+2A(x,y)\right)dx+\left(\frac{2y}{x^2+y^2}(2x+z)\right)dy+A(x,y)\,dz,$$
where $A\in C^1(\mathbb{R^2}\backslash(0,0); \mathbb{R})$
1) Find a function $A$ that makes the differential form exact.
2) Find all the primitives of $\omega$.
My idea was calculating $d\omega=0$ because $\omega$ is closed (necessary condition for exactness), and finding $A(x,y)$ who will depend on a costant $C$. And then, putting $\int_{\gamma}\omega=0$ in order to prove the exactness of $\omega$ for some curves $\gamma$.  
 A: Say that $f$ is a primitive of $\omega$, i.e. $\omega=df$. Since $A(x,y)$ is independent of $z$, we can find it by considering the pullback of $\omega$ to the $xy$-plane:
\begin{align}
\omega|_{z=0}&=\left(\frac{4x^{2}}{x^{2}+y^{2}}+2A(x,y)\right)dx+\left(\frac{4xy}{x^{2}+y^{2}}\right)dy\\
&=\frac{\partial f}{\partial x}(x,y,0)dx+\frac{\partial f}{\partial y}(x,y,0)dy.
\end{align} 
So we get
$$
\frac{\partial f}{\partial y}(x,y,0)=\frac{4xy}{x^{2}+y^{2}}=\frac{\partial}{\partial y}\left(2x\ln(x^{2}+y^{2})\right).
$$
So a good guess would be
$$
f(x,y,0)=2x\ln(x^{2}+y^{2}).
$$
Then
\begin{align}
\frac{\partial f}{\partial x}(x,y,0)&=\frac{4x^{2}}{x^{2}+y^{2}}+2\ln(x^{2}+y^{2})\\
&=\frac{4x^{2}}{x^{2}+y^{2}}+2A(x,y).
\end{align}
This makes us set
$$
A(x,y)=\ln(x^{2}+y^{2}).
$$
Now check that this guess for $A(x,y)$ works:
$$
\omega=\left(\frac{4x^{2}+2zx}{x^{2}+y^{2}}+2\ln(x^{2}+y^{2})\right)dx + \left(\frac{2y}{x^{2}+y^{2}}(2x+z)\right)dy+\ln(x^{2}+y^{2})dz
$$
is indeed exact with primitive
$$
f=(2x+z)\ln(x^{2}+y^{2}).
$$
So all primitives are of the form
$$
(2x+z)\ln(x^{2}+y^{2})+C
$$
for some constant $C$.
