at.algebraic topology - De Rham cohomology and antiderivatives

This probably doesn't answer your question, but maybe it is close.

The (Extended) Poincare lemma says that if U is a contractible smooth $n$-manifold and
$omega$ is a closed differentiable form meaning $domega=0$ then there is a form $phi$
with $dphi=omega$. So, $phi$ is an antiderivative of $omega$.

More to the point, if $omega$ is a closed form on the smooth manifold $M$ then $[omega]in H^*(M)$, and $omega$ has an antiderivative if and only if $[omega]=0$.

More specifically, suppose that $M$ is an oriented smooth $n$-manifold so that integration
is defined, and $omega$ is a smooth $n$-form, then $omega$ has an antiderivative if and only if $int_Momega=0$.

If $M$ is in addition compact, and $omega$ is a closed $i$-form then $omega$ has an antiderivative if and only if for all closed $n-i$ forms $eta$, $int_M omegawedge eta=0$.

If $omega$ is not closed it can never have an antiderivative.