Finding a function if you know it's directional derivative

I have the following expression $$$$\beta \,\,\cdotp \nabla v(x,y) = - u(x,y)$$$$ where $$\beta \,\,\cdotp \nabla$$ denotes the directional derivative of $$v$$ in the direction of $$\beta$$.

Question: How do I get to an explicit expression for $$v$$?

Intuitively I think I know what the solution should be. $$v$$ should be the line integral over $$u(x,y)$$ from $$(x,y)$$ to some constant point $$(a,b)$$, where the line along which $$u$$ is integrated is tangent to the vector $$\beta$$. In that case the directional derivative of $$v$$ in the direction of $$\beta$$ would be equal to $$-u(x,y)$$ because a small step in the direction of $$\beta$$ (or up along the line integral) would decrease the integral by $$-u(x,y)$$.

I don't know how to write this down though and I've looked for a lot of sources.

The given data do not determine the function $$(x,y)\mapsto v(x,y)$$ up to an additive constant. Consider the simple example $$v_x=u(x,y)\ ,$$ which corresponds to $$\beta=(-1,0)$$. In this case you might find a solution $$v_0$$ by "integrating $$u$$ with respect to $$x$$", but you can add an arbitrary function $$y\mapsto g(y)$$ to this $$v_0$$ and obtain another solution $$v(x,y):=v_0(x,y)+g(y)$$.