It's a little hard to distinguish between [itex]\nabla[/itex] and V!
What you are asking is how they went from
[tex]\int\int\int (\nabla\cdot E)V d\tau[/tex]
to
[tex]\int\int\int E\cdot\nabla V d\tau+ \int\int V (E\cdot dA)[/tex]
(some people prefer [itex]E\cdot n dA[/itex] rather than [itex]E\cdot dA[/itex] where "n" is the unit normal to the surface.) V here is the scalar potential and E is a vector function.
There are actually two steps in there. First they are using the "product" rule:
[tex]\nabla\cdot(VE)= V\nabla\cdot E+ (\nabla V)\cdot E[/itex]<br />
where [itex]\nabla\cdot E[/itex] and [itex]\nabla\cdot (VE)[/itex] are the divergence (div) of the vectors and [itex]\nabla V[/itex] is grad V.<br />
<br />
so<br />
[tex]\int\int\int\nabla\cdot(VE)d\tau= \int\int\int V\nabla\cdot Ed\tau +\int\int\int(\nabla V)\cdot E d\tau[/itex]<br />
<br />
Now use the divergence theorem to convert that first integral on the right to<br />
[tex]\int\int VE\cdot dA[/itex]<br />
<br />
But I'm not at all clear why the [itex]\epsilon_0/2[/itex] only multiplies the first integral![/tex][/tex][/tex]