Im trying to show that the integral over a closed loop of a crossproduct stays the same if I choose a different origin with $\overrightarrow{r}=\overrightarrow{r}\prime+\overrightarrow{r\_0}$ and $\oint{d\overrightarrow{F}}=0$

$$ \overrightarrow{N} = \oint{\overrightarrow{r} \times d\overrightarrow{F}} = \oint{(\overrightarrow{r}\prime+\overrightarrow{r\_0} )\times d\overrightarrow{F}} = \oint{\overrightarrow{r}\prime \times d\overrightarrow{F}} + \oint{\overrightarrow{r\_0} \times d\overrightarrow{F}}$$

How do I show $\oint{\overrightarrow{r\_0} \times d\overrightarrow{F}}= 0 $, or why can I say $\oint{\overrightarrow{r\_0} \times d\overrightarrow{F}} = \overrightarrow{r\_0} \times\oint{ d\overrightarrow{F}}$?

To clarify, $\overrightarrow{dF}$ is the force on the current $I \overrightarrow{dl}$ in a loop in a uniform and constant magnetic field. So $\overrightarrow{dF} = I \overrightarrow{dl} \times \overrightarrow{B}$.