$\nabla^{\mu}\nabla_{\mu}$ in general relativity I am trying to work out $\square=\nabla^{\mu}\nabla_{\mu}$ in the metric 
$
ds^{2}=-A(r)dt^{2}+B(r)^{-1}dr^{2}+r^{2}d\Omega^{2}
$$
My work:
when applying $\square$ to a scalar $\phi$, then
$
\square\phi=\nabla^{\mu}\nabla_{\nu}\phi=\nabla^{\mu}\partial_{\mu}\phi=g^{\mu\nu}\nabla_{\nu}\partial_{\mu}\phi=g^{\mu\nu}(\partial_{\nu}\partial_{\mu}-\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda})\phi
$
Christoffel symbol
\left(
\begin{array}{ccc}
 \left\{0,\frac{A'(r)}{2 A(r)},0\right\} & \left\{\frac{A'(r)}{2 A(r)},0,0\right\} & \{0,0,0\} \\
 \left\{\frac{1}{2} B(r) A'(r),0,0\right\} & \left\{0,-\frac{B'(r)}{2 B(r)},0\right\} & \{0,0,-r B(r)\} \\
 \{0,0,0\} & \left\{0,0,\frac{1}{r}\right\} & \left\{0,\frac{1}{r},0\right\}
\end{array}
\right)
substituting the metric and affine values in the equation above, my answer came to be 
$$
\square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{1}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr}
$$
However, the answer happens to be
$$
\square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{2}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr}
$$
Could someone please show me where the third comes from in the second term?
 A: I'll prove a formula that is probably easier to use for this.
\begin{equation}
\begin{split}
\frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} g^{\mu\nu} \partial_\nu \phi \right) &= \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g}  \right)  g^{\mu\nu} \partial_\nu \phi +  \partial_\mu  \left( g^{\mu\nu} \partial_\nu \phi \right)  \\
&=  \frac{1}{2g}  \partial_\mu g  g^{\mu\nu} \partial_\nu \phi +  \partial_\mu  g^{\mu\nu}   \partial_\nu \phi  +  g^{\mu\nu}  \partial_\mu \partial_\nu \phi  \\
&= - \frac{1}{2} g^{\alpha\mu}   g^{\nu\beta}  \left[ \partial_\mu   g_{\alpha\beta} +  \partial_\alpha g_{\mu\beta} -  \partial_\beta g_{\alpha\mu}  \right] \partial_\nu \phi  +  g^{\mu\nu}  \partial_\mu \partial_\nu \phi \\
&= g^{\mu\nu} \left( \partial_\mu \partial_\nu  - \Gamma^\lambda_{\mu\nu} \partial_\lambda \phi \right) \\
&= \nabla^\mu \nabla_\mu \phi 
\end{split}
\end{equation}
A: Mm.... At first I repeated your calculation and got the same answer as yours, then I checked the paper you gave and found it consist with (32 a) and it seems not a typo, so I read it from begining - oh brother it's not 2+1 gravity - -b it's 3+1 gravity and you should treat $d\Omega^2$ more carefully:
$r^2 d\Omega^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2$
so you get one more r factor when using the formula  @Prahar gives.
and btw the Christoffel symbol is for 2+1 and certainly wrong for 3+1. 
