I want to prove a statement: $$\nabla_{\mu}A^{\mu} = \frac{1}{\sqrt{-g}} \partial_{\mu}(\sqrt{-g} A^{\mu}).$$ I use the definition $\nabla_{\mu}A^{\nu} = \partial_{\mu}A^{\nu} + \Gamma^{\nu}_{\mu\lambda} A^{\lambda}$, so the initial expression becomes: $$\nabla_{\mu}A^{\mu} = \partial_{\mu}A^{\mu} + \Gamma^{\mu}_{\mu\lambda} A^{\lambda}$$ And I also know a property $\Gamma_{\nu\mu}^{\mu} = \partial_{\nu} \ln \sqrt{-g}$. So I try to $$\nabla_{\mu}A^{\mu} = \partial_{\mu}A^{\mu} + \frac{1}{\sqrt{-g}} \partial_{\mu}\big[\ln(\sqrt{-g}) A^{\mu}\big].$$ And that's where I'm stuck. Please, help me out, I want to understand how it's done. I can't get rid of the first term.
Upd: let's follow Winther's advice and expand the right-hand side: $$\frac{1}{\sqrt{-g}} \partial_{\mu}(\sqrt{-g} A^{\mu}) = \partial_{\mu}A^{\mu} + \frac{\partial_{\mu} g}{2g} A^{\mu} = \partial_{\mu}A^{\mu} + \frac{1}{2} \partial_{\mu} \ln g A^{\mu},$$ Q.E.D.