Find a function that makes this differential form exact We have $\Omega=\mathbb{R^3}\backslash \left\{ (0,0,z):z\in \mathbb{R}\right\}$ 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 $A\in C^1(\mathbb{R^2}\backslash(0,0); \mathbb{R})$
Find a function $A$ that makes the differential form exact.
My teacher advised to use the homotopy of the curves of which I make the integral. Namely find two curves on which the integral is $0$, one that surrounds the origin and the other doesn't, and hence obtaining all the other curves through continuous deformation. For the curves that don't include the $z$-axis I can just take a ball that contains the curve, and since the ball is convex, I know that a closed differential form in a open convex set is exact (I'm pretty sure it is closed). I'm stuck on the curves that surround the $z$-axis, any ideas?
 A: You must first verify that your form is at least closed, that is $d\omega = 0$. If I did everything well, after a lengthy calculation you'll get $A= \log(x^2+y^2)+C$. Finally check that $\int_C \omega = 0$ for some curve sourronding the $z$-axis, I chose the curve $C = (\cos t,\sin t, 0)$ for $0\le t\le 2\pi$, because the integral seems easy to evaluate. Since the interal is zero, your form is exact. 
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To answer the question of @JBond007, let me write two proofs.
(i) For a closed 1-form $\omega$ we are tempted to define the solution as $f(p)=c+\int_q^p\omega$, for some fixed point $q$. If two curves $C_1$ and $C_2$ join $q$ and $p$, and one of them can be deformed into the other, then they can be seen as the boundary of a surface $S$ and by Stokes theorem $\int_{C_1-C_2}\omega = \int_S d\omega = 0$. The problem here is that $M=\mathbb{R}^3-\{z \enspace\text{axis}\}$ is not simply-connected; however $C_1-C_2$ can be deformed into some loop $C = (\cos t,\sin t, 0)$, for $t\in[0,2\pi n]$, that turns $n$ times around the $z$ axis (smash $C_1-C_2$ into the plane $\{z=0\}$ and then contract to $S^1$), so by Stokes Theorem again $\int_{C_1-C_2}\omega=\int_C\omega$. Hence, if the integral around $C$ is zero, then there is no obstruction to define $f(p)=c+\int_q^p\omega$, since the definition does not depend on the curve joining $p$ and $q$.
(ii) You can use an exact sequence. Write $M=U_1\cup U_2$, where $U_1:=\mathbb{R}^3-\{x\ge 0 \text{ and } y=0\}$ and $U_2:=\mathbb{R}^3-\{x\le 0 \text{ and } y=0\}$. Both $U_1$ and $U_2$ are simply-connected, so you can find a solution $\omega = df_i$ in $U_i$. In $U_1\cap U_2$ we have that $0=df_1-df_2$, which implies that $f_1-f_2$ is constant in each connected component of $U_1\cap U_2 = \{y>0\}\cup\{y<0\}$, say $c_+$ in $\{y>0\}$ and $c_-$ in $\{y<0\}$. If $c_+=c_-=c$ then the solution $f = f_1$ in $U_1$ and $f = f_2+c$ in $U_2$ is well defined in $M$. The integral of $\omega=df_1$ on the half-loop $C_1=\{(\cos t,\sin t, 0)\mid t\in[\pi/2,3\pi/2]\}\subset U_1$ is $f_1(0,1,0)-f_1(0,-1,0)$, and the integral on the other half-loop inside $U_2$ is $f_2(0,-1,0)-f_2(0,1,0)$. Hence, if $\int_C\omega = 0$ then $0=(f_1(0,1,0)-f_2(0,1,0))-(f_1(0,-1,0)-f_2(0,-1,0))=c_+-c_-$, and $c_+=c_-$.
