What makes a Kähler manifold projective? Let $X$ be a compact Kähler manifold, I know there are (at least?) 2 ways to make $X$ a projective manifold.

*

*(integral condition) If the Kähler class $[\omega]$ is integral, i.e., $[\omega]\in H^2(X,\mathbb Z)$, then $X$ is projective.

*(Moishezon condition) If the Kähler manifold $X$ is also a Moishezon manifold, then $X$ is projective.

In summary, we may write:

*

*Kähler+integral=projective,

*Kähler+Moishezon=projective.

And my question is that: what's the relationship between integral condition and Moishezon condition?
It includes: which one is stronger? Or whether one can deduced from the other？ Or are they actually equivalent?
Can we unify these 2 conditions together to get only one single condition? Or can we abstract the essence of both these 2 conditions?
I guess the Moishezon condition is another version of "integral" condition, this view comes from a post from @Gunnar Þór Magnússon

A Moishezon manifold can now be characterized as a modification of a projective manifold, so there exists a generically 1-1 meromorphic map $X\to Y$, where $Y$ is projective. This is equivalent to $X$ admitting an integral Kähler current, which is roughly a closed (1,1)-form $T$ that has distribution coefficients (in local coordinates) whose cohomology class is integral.

So can we unify these 2 conditions together to just one condition: $X$ admits an integral Kähler current?
 Let me repeat my question again: what's the relationship between integral condition and Moishezon condition?
Added later:
I know $X$ admits an integral Kähler class if and only if $X$ has a positive line bundle, later I learned a compact complex manifold is Moishezon if and only if it carries a big line bundle. So in the view of "line bundles", we may write:

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*Kähler+positive=projective

*Kähler+big=projective

And then it turns to the relationship between big and positive.
So what's the essence that makes a compact Kähler manifold projective? I apologize that my question seems a little bit vague, so any comment clarifying it is welcome!
 A: If I understand the question correctly, I think that the answer is given by the main result in
S. Ji: Currents, metrics and Moishezon manifolds, Pac. J. Math. 158, No. 2, 335-351 (1993). ZBL0785.32011.
Essentially, the existence of a bimeromorphic modification that is projective gives a $d$-closed (1, 1)-current on our manifold (the pushforward of the Kähler form on the modification) that satisfies three properties. Conversely, the existence of such a current implies that the manifold is Moishezon.
Here is the rewriting in LaTex of the first page (warning: there is a misprint in the second line of the Introduction paragraph, the projective algebraic manifold is $\widetilde{M}$, not $M$).

Abstract:  A compact complex manifold $M$ is Moishezon if and only if there exists an integral closed positive $(1, 1)$-current $\omega$ such that $\omega \ge \epsilon\sigma$ and $\omega$ is smooth outside an analytic subvariety.
1.  Introduction.  Given a Moishezon manifold $M$, it is well known (cf. [Mo], [W]) that there is a bimeromorphic morphism $\pi : \widetilde M \to M$ such that the manifold $M$ is projective algebraic.  Let $\tilde\omega$ be Kähler form on $\widetilde M$ with $[\tilde\omega] \in H^2(\widetilde M, \mathbf Z)$.  Then the pushforward current $\omega = \pi_*\tilde\omega$ is a $d$-closed current on $M$ such that
(i) $\omega \in H^2(M, \mathbf Z)$;
(ii) $\omega$ is smooth on $M - S$, where $S$ is some proper analytic subset in $M$;
(iii) $\omega \ge \epsilon\sigma$ in the sense of currents, where $\epsilon > 0$ is some real number and $\sigma$ is a fixed positive definite $(1, 1)$-form (not necessarily $d$-closed) on $M$.
Conversely, we shall prove the following
Theorem 1.1.  Let $M$ be a compact complex manifold of dimension $n$.  Then $M$ is Moishezon if and only if there exists a $d$-closed $(1, 1)$-current $\omega$ on $M$ such that the conditions (i), (ii) and (iii) above are satisfied.

