For a commutative algebra $A$, let a biderivation $P$ be called a Poisson structure if $[[P,P]]=0$ (the bracket is Schouten-Nijenhuis). Then one obtains a complex of multiderivations with $[[P,{}]]$ being the differential, and so one gets Poisson cohomology. Due to graded Jacobi identity for Schouten-Nijenhuis bracket, Poisson cohomology inherits this bracket.
Let $P$ be called nondegenerate if $da \mapsto P(a,{})$ defines an isomorphism between 1-forms and 1-derivations. (Here one also needs to assume that the module of 1-forms is projective and finitely generated.)
How can one see that for a nondegenerate Poisson structure Schouten-Nijenhuis bracket on cohomology is trivial?
I highly suspect that for X and Y from cohomology we have $[[X,Y]]=[[P,i_\Omega(X\wedge Y)]]$ where $\Omega$ is the symplectic 2-form corresponding to $P$ and $i$ is insertion, but I only managed to show that for $|X|=1$ and $|Y|=0$ which is a little bit not enough.
I've done it myself but the full proof is a little bit long so here is the pdf file: https://www.dropbox.com/s/d368qulnyzn82v4/Poisson.pdf?dl=0.
The idea is that there exists a bracket on differential forms that is trivial on de Rham cohomology for any Poisson structure and that coincides with Schouten-Nijenhuis bracket if Poisson structure is nondegenerate.