How Is the Divergence of J Derived in Electromagnetic Fields?

[barefeet]

Homework Statement

In a book I find the following derivation:
$$\int (J \cdot \nabla ) \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'}= -\sum_{i=1}^3 \int J_i \frac{\partial}{\partial r_i'} \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'} \\ = -\sum_{i=1}^3 \int J_i \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^2r_{j\neq i}'\bigg|_{r_i' = -\infty}^\infty + \int \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} \nabla' \cdot \mathbf{J} d^3\mathbf{r'}$$

Homework Equations

None

The Attempt at a Solution

I understand the first line, but in the second line I don't understand how the second term comes about. I don't see how the divergence of J is taken as the nabla operator doesn't operate on it. If I calculate for example one of the three terms:
$$\int J_x \frac{\partial}{\partial x'} \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'}$$
I don't see how a term with $\frac{\partial J_x}{\partial x'}$ comes out of this?

 

[jasonRF]

Here is a hint: integration by parts

 

[cpsinkule]

the x component of j*grad (r/r³) is j*grad(r_x/r³), not j_xgrad(r_x/r³) you then use the vector calculus identity
div(fV)=fdivV+V*grad(f)