# How to find a form whose exterior derivative is an exact differential form?

I think it will help me if look at examples. So, in other words, let's say we have an exact differential form:

$$\alpha = (yz-z)dx + (xz+z)dy + (xy-x+y)dz$$

How can I find a differential form $\omega$ such that the exterior derivative of $\omega$ is $\alpha$? Is it just integration? And if so, how would that look? I don't yet have a good understanding of integrating differential forms.

And is it a different or more difficult process if we have an exact 2-form?

For instance, say we have the 2-form:

$$\beta = 2xy^2 dxdy +z dydz$$

Thanks very much!

For the second example, if you know the basic rules for $d$ you can just guess an answer because there's no intermixing of variables as there is in the first. If you take $d(x^2y^2\,dy)$, you'll get the first term and if you take $d(yz\,dz)$, you'll get the second. (There are, of course, other possibilities.)
For the first example, this is just the procedure you should have learned in multivariable calculus to find the potential function for a conservative vector field. You want a function $f(x,y,z)$ with \begin{align*} \frac{\partial f}{\partial x} &= yz-z \\ \frac{\partial f}{\partial y} &= xz+z \\ \frac{\partial f}{\partial z} &= xy-x+y \end{align*} You integrate the first with respect to $x$ to find $f(x,y,z) = xyz - xz + g(y,z)$ for some (so far) unknown function $g$. So then the second tells you that $\dfrac{\partial g}{\partial y} = z$, so $g(y,z) = yz + h(z)$ for some unknown $h$. The third now tells us that $h'(z) = 0$, so $h(z)=c$ for some constant. So we have $f(x,y,z) = xyz-xz+yz+c$. Check by taking $df$.