electrostatics - Direction of $dmathbf{l}$

A solid sphere has charge $q$ and radius $R$. Find the potential at a point a distance $r$ from the center of the sphere where $r>R$, using infinity as the reference point.

My attempt:

From Gauss' theorem we may deduce that $\displaystyle\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r'^2}\hat{\mathbf{r}}$ where $r'$ is the distance of an arbitrary point from the center of the sphere provided $r'>R$.

$V=-\int_\infty^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$

My question: What is $d\mathbf{l}$? Since we are traversing in the direction opposite to $\hat{\mathbf{r}}$, I think it should be $-dr'\hat{\mathbf{r}}$. But when I use it to find $V$, I get a sign error. Please help!

Answer

I am going to contradict the answers by @Bio (whose answer has since been deleted) and @lineage and say that $\text d\mathbf l$ is actually $\text d r'\hat{r}$ The other answers are mathematically correct, but it goes against our physical intuition with how the limits of integration are set up, as it seems you were discussing in the comments of the answer.

Indeed, it would be nice if our lower limit of integration was where we started and the upper limit was where we ended. This makes so much more sense if we make $\text d\mathbf l=\text d r'\hat{r}$. This is because the sign of $\text dr'$ is actually determined by our limits already. In general, if we are integrating from $r'=a$ to $r'=b$ we will have $$\Delta V=-\int_a^bE_r\text dr'$$

If $b>a$ then $\text dr'$ is positive, and if $b (which is what you are looking at) then $\text dr'$ is negative. The sign is already taken care of depending on how you set up the limits! You don't need to explicitly put in the sign of $\text dr'$

This is why in @Bio's answer (as well as @lineage's answer I believe, although that answer is very convoluted, so I am unsure) you need to switch the limits of integration. That way you are doing two sign changes, resulting in the same integral. While this is mathematically correct, I feel like you really lose the physical intuition of adding up these values as you move from the start to the end of the path. @Bio's integral is technically moving backwards along the path while adding up the negative of the values given by the integrand, thus yielding the same result.