$\omega = a\arctan(xe^y)dx+e^{-y}\ln(1+x^2e^{2y})dy$
Find $a$ s.t. $\omega$ is exact. Find a primitive of $\omega$.
I'm very confused about differential forms, so I don't know how to approach this question. To find $a$, I tried using the lemma that we saw in class i.e. if $ \frac{\partial \alpha_i}{\partial x_j}=\frac{\partial \alpha_j}{\partial x_i}$ then $\omega$ is exact. So I calculated $\frac{\partial}{\partial x_2}a\arctan(xe^y)$ and $\frac{\partial}{\partial x_1}e^{-y}\ln(1+x^2e^{2y}) $ and that got me to the conclusion that $a=2$.
Now for the primitive, we only saw one example in class and it wasn't a very helpful one. So far I've integrated $2\arctan(xe^y)$for x. But what about $e^{-y}\ln(1+x^2e^{2y})$? What do I do with that? The idea would be to have this equation and find C: $\frac{\partial f}{\partial x_2} + C'(y)=e^{-y}\ln(1+x^2e^{2y})$ but it gets weird... Any idea of what I'm doing wrong?
Thanks!
Turns out I was doing everything right. So here are the details:
If $\frac{\partial f}{\partial x}=2\arctan(xe^y)$, then $f$ is the primitive of $2\arctan(xe^y)$ = $2x\arctan(xe^y)-e^{-y}\ln(x^2e^{2x}+1)+C(y)$
Then you figure out C by deriving f. So you would have $\frac{\partial f}{\partial y}(2x\arctan(xe^y)-e^{-y}\ln(x^2e^{2x}+1)+C(y))=e^{-y}\ln(x^2e^{2x}+1)+C'(y)$
So that brings us to the conclusion that $C'(y)=0$ i.e. it's a constant.
So the final answer is $f=2x\arctan(xe^y)-e^{-y}\ln(x^2e^{2x}+1)+C$