Equation of motion from Polyakov action

If I write (my understanding of) how to derive the equations of motion from the Polyakov action, I come up missing a term.

Beginning with the basic Polyakov action
begin{equation}
S = -frac{T}{2}intsqrt{-h}h^{ab}g_{munu}partial_aX^mupartial_bX^nu,d^2xi
end{equation}
the Lagrangian is
begin{equation}
mathcal{L}=-frac{T}{2}sqrt{-h}h^{ab}partial_aX^mupartial_bX^nu
end{equation}
which after plugging into the Euler-Lagrange equation is
begin{equation}
0-frac{partial}{partial xi^b}left[-frac{T}{2} sqrt{-h}h^{ab}partial_aX^nuright]=0
end{equation}
which leads to (my result of)
begin{equation}
partial_bleft[ sqrt{-h}h^{ab}partial_a X^nuright] =0
end{equation}
The equation of motion should be
begin{equation}
Box X^nu = frac{1}{sqrt{-h}}partial_bleft[ sqrt{-h}h^{ab}partial_a X^nuright] =0
end{equation}
My question is that I do not see where the $1/sqrt{-h}$ comes from. (This result is ok for the Polyakov action on its own, but I want to be able to add a mass term to the action which will make the $1/sqrt{-h}$ matter then.)

I think that there is something fundamental that I am missing about deriving the equation of motion. I understand that the operator $nabla_mu X^nu=partial_mu X^nu +Gamma^nu_{mulambda}X^lambda$, but I am bothered by deriving the equations of motion when the Lagrangian density is $L(X^nu,partial_aX^nu)$ instead of $L(X^nu,nabla_aX^nu)$. If I derive the Euler-Lagrange equation of motion from varying an action containing $L(X^nu,partial_aX^nu)$:
begin{eqnarray}
delta S &=& int delta L (X^nu,partial_a X^nu)sqrt{-g}d^nX \
&=& int left[ frac{delta L}{delta(partial_aX^nu)}delta(partial_aX^nu) + delta X^nunabla_afrac{delta L}{delta(partial_aX^nu)}right]sqrt{-g}d^nX + int left[ frac{delta L}{delta X^nu}-nabla_afrac{delta L}{delta (partial_a X^nu)}right]delta X^nusqrt{-g}d^nX
end{eqnarray}
can I still send the surface term (the first term above) to zero? Thanks for any insight.