$\textbf{b}:\Omega\to\mathbb{R}^2$ is a $C^1$ vector field. Need to prove the following version of the Green's formula: $$(\nabla u,\textbf{b}v)=-(u\textbf{b},\nabla v)-(u,v\nabla\cdot \textbf{b})$$ where $(\cdot,\cdot)$ is the $L_2$ inner product.
I started from the left side $(\nabla u,\textbf{b}v)=\int_\Omega\nabla u\cdot\textbf{b}v$. My plan was to get to $$-\int_\Omega u\textbf{b}\cdot\nabla v-\int_{\partial\Omega}uv\nabla\cdot\textbf{b}$$ using integration by parts. But I am having troubles coming up with logical steps. Also I am not sure about the domain of the second integral above.
Consider the following vector identity \begin{align} \text{div} (u (\mathbf{b}v))&= u \text{div} (\mathbf{b}v)+\nabla u \cdot \mathbf{b}v \\ &= u (\text{div}(\mathbf{b})v+\mathbf{b}\cdot\nabla v)+\nabla u \cdot \mathbf{b}v \\ &=u \text{div}(\mathbf{b})v+u\mathbf{b}\cdot\nabla v+\nabla u \cdot \mathbf{b}v \end{align} by Divergence Theorem \begin{align} \int_{\partial \Omega} uv (\mathbf{b} \cdot \mathbf{n})~dS = (uv,\text{div}~\mathbf{b})+(u\mathbf{b},\nabla v)+(\nabla u,\mathbf{b}v) \end{align} For smooth functions with compact support the left-hand side is zero and the conclusion follows.