If $E(L)$ is the Euler-Lagrange of $L$ and $$ L(t,x^i,\dot{x}^i) = B(t,x^i,\dot{x}^i) + A_j(x^i)\dot{x}^j $$ and $$ \frac{\partial A_j}{\partial x^j} = \frac{\partial A_j}{\partial x^i} $$ show that $$ E(L) = E(B) $$ for $n=1$.
Firstly, am I right to assume from the above that $ x^i = x^j $ ?
Secondly, if I show that $ E(A_j(x^i)\dot{x}^j) = 0 $ then I will have success?
$$ \frac{d}{dt}\left(\frac{A_j(x^i)\dot{x}^j }{\partial \dot{x}^i} \right) - \frac{\partial A_j(x^i)\dot{x}^j}{\partial x^i} = \frac{d}{dt}\left( \frac{\partial A_j}{\partial \dot{x}^i}\dot{x}^j + A_j\frac{\partial \dot{x}^j}{\partial \dot{x}^i} \right) - \frac{\partial A_j}{\partial x^i}\dot{x}^j - A_j\frac{\partial \dot{x}^j}{\partial x^i} $$
$A_j$ is a function of $x^i$ only, so $\frac{\partial A_j}{\partial \dot{x}^i}\dot{x}^j = 0$. So if $ x^i = x^j $, then $$ \frac{d}{dt}\left( \frac{\partial A_j}{\partial \dot{x}^i}\dot{x}^j + A_j\frac{\partial \dot{x}^j}{\partial \dot{x}^i} \right) - \frac{\partial A_j}{\partial x^i}\dot{x}^j - A_j\frac{\partial \dot{x}^j}{\partial x^i} = \frac{dA_j}{dx^i}\dot{x}^i- \frac{d A_j}{d x^i}\dot{x}^i - A_j\frac{\partial \dot{x}^i}{\partial x^i} = - A_j\frac{\partial \dot{x}^i}{\partial x^i} $$
I am missing something fundamental here!
Appreciate your help.
Hints:
We are given a closed 1-form $$A~=~\sum_{i=1}^n A_i \mathrm{d} x^i, \qquad \mathrm{d}A~=~0. \tag{1}$$
From Poincare Lemma, we know that locally in a neighborhood $U\subseteq \mathbb{R}^n$ there exists a 0-form/function $f$ such that $$A|_{U}~=~ \mathrm{d}f.\tag{2} $$
The Euler-Lagrange operator $E(L)$ in a point only depends on local data so we may w.l.o.g. replace $A_i$ with $\frac{\partial f}{\partial x^i}$.
Then the term $$\sum_{i=1}^n A_i \frac{dx^i}{dt}~=~\frac{df}{dt}\tag{3}$$ in the Lagrangian $L$ is just a total time derivative.
It is a general fact that the Euler-Lagrange operator $E(L)$ does not depend on total time derivative terms in the Lagrangian $L$.