The use of the formula $\displaystyle V(r) - V(\infty) = -\int^r_{\infty} \vec{E}\cdot d\vec{\ell}$ requires a little care.
If there is an electric field $\vec E$ then the force on unit positive charge due to this electric field is $\vec E$.
The work done by the electric field in moving this charge $d \vec r$ is $\vec E \cdot d\vec r$ and the work done by an external force is $\left(- \vec E \right) \cdot d\vec r$.
There are two equivalent ways to define the potential at a point.
The first definition is that the potential at a point is minus the work done by the electric field in taking unit positive charge from infinity to the point.
The work done by the electric field in taking unit positive charge from infinity to the point is $\displaystyle \int_\infty ^r \vec E \cdot d\vec r$
The potential at point $r$ is $V_r - V_\infty = V_r = \displaystyle \int_\infty ^r \vec E \cdot d\vec r$
Your uncertainty seems to come from what to do with $\vec E \cdot d\vec r$.
Since the path is radial there will be no angular dependence and so the dot product will be $E\; dr$ where $E$ is positive if the electric field is radially outwards or negative if the electric field is radially inwards.
The $dr$ is just a step length and its direction is taken care of by the limits of integration.
On working out $\displaystyle \int_{\infty}^{r} \dfrac {kq}{r^2}\;dr$ you get $- \dfrac {kq}{r}$ as the work done by the electric field.
If the path had been in the opposite direction ie from $r$ to $\infty$ the work done by the electric field would have been $+\dfrac {kq}{r}$.
This is exactly what you should expect.
With the first integral the force is outwards (positive direction) and the path is inwards (negative direction) and so the dot product will be negative.
With the second integral the force is outwards (positive direction) and the path is outwards (positive direction) and so the dot product will be positive.
To finish off remembering that the potential at a point is minus the work done by the electric field in taking unit positive charge from infinity to the point we get
$V_r – V_\infty = V_r = - \left ( - \dfrac {kq}{r } \right ) = + \dfrac {kq}{r } $
So in your original equation $\displaystyle V(r) - V(\infty) = -\int^r_{\infty} \vec{E}\cdot d\vec{\ell}$ is minus the work done by electric field.
The second definition is that the potential at a point is the work done by an external force in taking unit positive charge from infinity to the point.
For the external force to move the unit positive charge this force must equal $-\vec E$ and so the work done by the external force in taking unit positive charge from infinity to the point is
$\displaystyle \int_\infty ^r \left(- \vec E \right)\cdot d\vec r$
which is the same as
$\displaystyle - \int_\infty ^r \vec E \cdot d\vec r = $ minus the work done by the electric field.
Doing the integration remembering that the external force is $-\vec E$ gives
$\displaystyle \int_\infty ^r \left(- \dfrac {kq}{r^2}\right) dr = + \dfrac {kq}{r} = V_r$ as before.
So rewriting you original equation slightly $\displaystyle V(r) - V(\infty) = \int^r_{\infty} \left(-\vec{E}\right)\cdot d\vec{\ell}$ is the work done by the external force.
Best Answer
I am not sure but I think: $V=\int_r^{\infty}\vec{E}d\vec{r}$, that is why it is positive.