Is $\frac{\nabla{u(\mathbf{x})}}{(\nabla{u(\mathbf{x})})\cdot\mathbf{x}}$ the gradient of any function? Suppose 
$$\mathbf{p}(\mathbf{x})=\frac{\nabla{u(\mathbf{x})}}{(\nabla{u(\mathbf{x})})\cdot\mathbf{x}},$$
where $\mathbf{x} \in \mathbb{R}^n$, $\mathbf{p}:  \mathbb{R}^n \to \mathbb{R}^n$ and $u:\mathbb{R}^n \to \mathbb{R}$. Assume $\mathbf{p}$ and $u$ are convex and defined and differentiable for all $\mathbb{R}^n > \mathbf{0}$.
I would like to show that $\mathbf{p}(\mathbf{x})$ is the gradient of some function $B: \mathbb{R}^n \to \mathbb{R}$, although I don't necessarily need to know the form of the function. What I really need to do is calculate $B(\mathbf{x_1})-B(\mathbf{x_0})$ for arbitrary values of $\mathbf{x_1}$ and $\mathbf{x_0}$. 
I can already calculate $\mathbf{p}(\mathbf{x})$ as needed, and if I can show that it is a valid gradient, I think the gradient theorem says that $B(\mathbf{x_1})-B(\mathbf{x_0})$ = $\int^\mathbf{x_1}_\mathbf{x_0}{\mathbf{p}(\mathbf{x}) \cdot d\mathbf{x}}$, which I can calculate numerically. However, if $\mathbf{p}(\mathbf{x})$ is not a valid gradient of any function, then the line integral is path-dependent and $B$ is ill-defined.
So my question is, is it possible to show that $\mathbf{p}(\mathbf{x})$ as defined here is a valid gradient? Or are there conditions that I could apply to guarantee that it is? I've spent a couple of days trying out variations on the chain rule, substituting terms, etc., but haven't made much headway.
 A: $\newcommand{\dd}{\partial}\newcommand{\Del}{\nabla}\newcommand{\Vec}[1]{\mathbf{#1}}$Denoting the component functions of $\Vec{p}$ by $(p_{1}, \dots, p_{n})$, and assuming the function $u$ is of class $C^{2}$, the necessary integrability condition
$$
\frac{\dd p_{j}}{\dd x_{i}} = \frac{\dd p_{i}}{\dd x_{j}}\quad\text{for all $i$, $j$,}
\tag{*}
$$
encodes equality of mixed partial derivatives for the (prospective) potential function $B$ of $\Vec{p}$. If the domain of $u$ is simply-connected, this integrability condition is also sufficient. (In three variables, this is the "vanishing curl" condition $\Del \times \Vec{p} = 0$.)
In your situation (omitting dependence of the gradient on $\Vec{x}$ out of laziness),
$$
p_{i} = \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd u}{\dd x_{i}}.
$$
Consequently,
\begin{align*}
\frac{\dd p_{i}}{\dd x_{j}}
&= \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd^{2} u}{\dd x_{j}\, \dd x_{i}}
 - \frac{1}{(\Del u \cdot \Vec{x})^{2}}\, \frac{\dd u}{\dd x_{i}} \, \frac{\dd}{\dd x_{j}}(\Del u \cdot \Vec{x}) \\
%
&=  \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd^{2} u}{\dd x_{j}\, \dd x_{i}}
 - \frac{1}{(\Del u \cdot \Vec{x})^{2}}\, \frac{\dd u}{\dd x_{i}} \left[\frac{\dd u}{\dd x_{j}} + \Del\left(\frac{\dd u}{\dd x_{j}}\right) \cdot \Vec{x}\right] \\
%
&=  \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd^{2} u}{\dd x_{j}\, \dd x_{i}}
 - \frac{1}{(\Del u \cdot \Vec{x})^{2}}\, \frac{\dd u}{\dd x_{i}} \left[\frac{\dd u}{\dd x_{j}} + \sum_{k} x_{k} \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{j}}\right].
\end{align*}
That is, $\Vec{p}$ is a gradient field if and only if
$$
\frac{\dd u}{\dd x_{i}} \sum_{k} x_{k}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{j}}
= \frac{\dd u}{\dd x_{j}} \sum_{k} x_{k}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{i}}\quad\text{for all $i$, $j$,}
$$
if and only if
$$
\sum_{k} x_{k} \left[\frac{\dd u}{\dd x_{i}}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{j}} - \frac{\dd u}{\dd x_{j}}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{i}}\right] = 0\quad\text{for all $i$, $j$.}
\tag{**}
$$
Equation (*) does not generally hold for your $\Vec{p}$. For instance, if $u(x, y) = x + y^{2}$, then $\Del u = (1, 2y)$ and $\Del u \cdot \Vec{x} = x + 2y^{2}$, so
$$
\Vec{p} = \frac{(1, 2y)}{x + 2y^{2}};\qquad
p_{1} = \frac{1}{x + 2y^{2}},\quad
p_{2} = \frac{2y}{x + 2y^{2}}.
$$
A short computation gives
$$
\frac{\dd p_{1}}{\dd y} = -\frac{4y}{(x + 2y^{2})^{2}}
\neq -\frac{2y}{(x + 2y^{2})^{2}} = \frac{\dd p_{2}}{\dd x}.
$$
In any event, and (*) or (**) is easier to check than the integral condition you've been considering.
