Can a vector be integrated with respect to another vector?

As janmarqz asked, what type of vector product do you intend? There are two ways to multiply two vectors- dot product and cross product. If you mean dot product then, writing $\vec{v}= <x, y, z>$ and $d\vec{v}= <dx, dy, dz>$, then $\int \vec{v}\cdot d\vec{v}= \int xdx+ ydy+ zdz= \frac{1}{2}(x^2+ y^2+ z^2)+ C$.

If you mean the cross product, then $\vec{v}\times d\vec{v}= (ydz- zdy)\vec{i}+ (zdx- xdz)\vec{j}+ (xdy- ydx)\vec{k}$. Since the integrals of each component is yz- zy= 0, zx- xz= 0, and xy- yx= 0, $\int \vec{v}\times d\vec{v}= \vec{0}$, the 0 vector. (More simply, $\vec{v}$ and $d\vec{v}$ are vectors in the same direction so their cross product is the 0 vector.)

As said by others in the comments, to answer this question we must first specify what kind of product is intended between the differential $d\vec{v}$ and the integrand. If the product we have in mind is the dot product, then this integral indeed exists.

Let $\vec{F}:\mathbb R^n\to\mathbb R^n$ be a vector field. Then the integral $$\int_\gamma\vec{F}(\vec{v})\cdot d\vec{v}$$ where $\gamma$ is a path in $\mathbb R^n$, can be defined. If we interpret $\gamma$ as the path of a particle and we interpret $\vec{F}$ as a force, then an intuitive interpretation of this integral is as the continuous sum of forces acting on the particle as it travels along its path. This is a physical quantity known as work, which is extremely useful in physics.