Is true that a closed $k$-form defined on an open connected set is exact iff its integral on a compact $k$-manifold without boundary is zero? So let be $U$ an open connected set of $\Bbb R^n$ and thus let be $\omega$ a $k$-form there defined. So if $\omega$ is exact than by the Stoke's theorem
$$
\int_M\omega=0
$$
where $M$ is a compact $k$-manifold contained in $U$. However if $\omega$ is such that the last identity holds for any compact $k$-manifold contained in $U$ then is it exact? How prove it? Is it much complicate? What I have to study to prove it? Where can I find the proof? So could someone help me, please?
 A: This is true but as far as I know it is a hard theorem without any elementary proof.  By the de Rham theorem, $\omega$ is exact iff its integral over every smooth $k$-cycle is $0$.  So, your question is equivalent to asking whether the homology group $H_k(U;\mathbb{R})$ is generated by the fundamental classes of closed oriented submanifolds $M\subseteq U$.  This (stated more generally for arbitrary manifolds) is Corollary II.30 of Thom's famous paper

Thom, René, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28, 17-86 (1954). ZBL0057.15502.

which introduced cobordism theory and earned him his Fields medal.
To very briefly sketch the argument, Thom uses transversality theory to show that if $f:U\to MSO(m)$ is any continuous map that approaches the basepoint as you approach the boundary of $U$, and $u\in H^m(MSO(m))$ is the Thom class, then the Poincaré dual of $f^*(u)\in H_c^m(U)$ is the fundamental class of a closed oriented submanifold of $U$ (namely, the inverse image of the zero section in $MSO(m)$ when you homotope $f$ to be transverse to it).  So, it suffices to show that $H_c^m(U;\mathbb{R})$ is generated by classes obtained by pulling back the Thom class along such maps $U\to MSO(m)$.  Every class in $H_c^m(U)$ is classified by a map from $U$ to the $n$-skeleton of $K(\mathbb{Z},m)$.  By induction on $n$, you can show that the universal $m$th cohomology class on the $n$-skeleton of $K(\mathbb{Z},m)$ has a multiple that is pulled back from the Thom class via a map to $MSO(m)$ (since the relevant obstruction group for extending from the $n$-skeleton to the $(n+1)$-skeleton is torsion).  Thus every class in $H_c^m(U)$ has a multiple which is Poincaré dual to the fundamental class of a closed oriented submanifold, and so $H_c^m(U;\mathbb{R})$ is generated by such classes.
