In the variational calculus, if the Lagrangian $L[y(x),y'(x)]$ is not an explicit function of $x$, the Euler-Lagrange equation takes on the following form:
$$\frac{d}{dx}\left(L-y'\frac{\partial L}{\partial y'}\right)=0\,.$$
This can be integrated to arrive at the Beltrami identity. However, if the Lagrangian depends explicitly on $x$, one has to use the full Euler-Lagrange equation. In this case, it can be recast in a form that is similar to the above, but with nonzero right-hand side:
$$\frac{d}{dx}\left(L-y'\frac{\partial L}{\partial y'}\right)=\frac{\partial L}{\partial x}\,.$$
I would like to integrate this equation but am not sure how to do it when the left-hand-side contains the total derivative and the right-hand side only a partial derivative. Is there any general solution to this problem, similarly as the Beltrami identity "undoes" the derivative in the first equation above?