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It might just be a typo, but you did not write out integration over $c$ in the last equation. We should have $${\partial s \over \partial t}+\text{div}(j_s)+k\int{dc\bigg({\partial f \over \partial t}+\sum_{i}{c_i{\partial f \over \partial r_i}}\bigg)}=k[\ln{f_0}+\Phi,\hat{\Omega}\Phi]$$ Now we come back to your question. Since you are reading the paper of your link, you may notice there is this property about the bracket integral $$[X,\hat{\Omega}Y]=[Y,\hat{\Omega}X]$$ In addition, if some function $X$ is independent of the velocity $c$, the term $\hat{\Omega}X$ is $0$. This can be seen from the explicit form of collision integral on page 13 and 14 of the paper, just around Eq. 2.1. In physical explanation, the property is easy to understand since the collision integral depicts the effect of momentum exchange of particles before and after collisions with the function $X$, if $X$ is independent of $c$, there is no momentum exchange to include. This means, \begin{split} \int{dc\bigg({\partial f \over \partial t}+\sum_{i}{c_i{\partial f \over \partial r_i}}\bigg)} & = \int{dc I(f)} = \int{dc \hat{\Omega}\Phi} \\ & = [1,\hat{\Omega}\Phi] = [\Phi,\hat{\Omega}1] = 0 \end{split} And we have similarly for $[\ln{f_0},\hat{\Omega}\Phi]=[\Phi,\hat{\Omega}\ln{f_0}]=0$. Hope this answers your question.