I find it easier to think about this by recognizing that
-\mu\nabla^2\mathbf{U} - (\mu + \lambda)\nabla(\nabla\cdot\mathbf{U}) = -\nabla\cdot\pmb{\sigma}\text{ ,}
where
\pmb{\sigma} = \lambda(\nabla\cdot\mathbf{U})\mathbf{I} + 2\mu\operatorname{sym}(\nabla\mathbf{U})\text{ .}
Then you can use the following integration-by-parts identity:
-\int_\Omega(\nabla\cdot\pmb{\sigma})\cdot\mathbf{v}\,d\mathbf{x} = \int_\Omega \pmb{\sigma}:\nabla\mathbf{v}\,d\mathbf{x} - \int_{\partial\Omega}(\pmb{\sigma}\mathbf{n})\cdot\mathbf{v}\,d\mathbf{s}\text{ ,}
where the boundary term is often left out, to weakly enforce the traction-free boundary condition, \pmb{\sigma}\mathbf{n} = \mathbf{0}.
If that integration-by-parts identity is confusing, I wrote up a series of exercises to guide students through its derivation, which can be found in an appendix of the “mathematical background” notes from my graduate course (although, for those stumbling on this link in the far future, I will likely rearrange that page over subsequent iterations of the course, starting in a month or two).