I saw this Lagrangian in notes I have printed:
$$ L\left(x,\frac{dx}{dt}\right) = \frac{m^2}{12}\left(\frac{dx}{dt}\right)^4 + m\left(\frac{dx}{dt}\right)^2 V(x) -V^2(x). $$ [It appears in Chapter 1 of Goldstein's Classical Mechanics as Exercise 18 (20) in the 2nd (3rd) edition, respectively.]
What is it? Is it even physical? It seems like it doesn't have the right units of energy.
Lagrangian in 1D: $$L~=~\frac{1}{3}T^2+2TV-V^2, \qquad T~:=~\frac{m}{2}\dot{x}^2. \tag{L}\label{eq:L}$$
Euler-Lagrange equation: $$\begin{align}2(T-V)V^{\prime} ~=~&\frac{\partial L}{\partial x} ~=~ \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial L}{\partial \dot{x}}\right)\cr ~=~& \frac{\mathrm{d}}{\mathrm{d}t} \left[\left(\frac{2}{3}T +2V\right)m\dot{x}\right] \cr ~=~& \left(\frac{2}{3}T +2V\right)m\ddot{x} + \left(\frac{2}{3}m\dot{x}\ddot{x} +2V^{\prime}\dot{x}\right)m\dot{x}\cr ~=~& 2(T+V)m\ddot{x} +4TV^{\prime}, \end{align}\tag{EL1}\label{eq:EL1}$$ or equivalently $$ 2(T+V)(m\ddot{x}+V^{\prime})~=~0. \tag{EL2}\label{eq:EL2}$$
Eq. $\eqref{eq:EL2}$ implies either (i) Newton's 2nd law $$ m\ddot{x}~=~-V^{\prime}, \tag{N2}\label{eq:N2}$$ or (ii) zero mechanical energy $$ T+V~=~0. \tag{E0}\label{eq:E0}$$
Zero mechanical energy $\eqref{eq:E0}$ is a special case of constant mechanical energy $$ T+V~=~E_0. \tag{E}\label{eq:E}$$
Constant mechanical energy $\eqref{eq:E}$ is equivalent to $$ 0~=~\frac{\mathrm{d}(T+V)}{\mathrm{d}t}~=~\dot{x}(m\ddot{x}+V^{\prime}). \tag{E'}\label{eq:E'}$$
Eq. $\eqref{eq:E'}$ in turn implies either Newton's 2nd law $\eqref{eq:N2}$ or zero velocity $$ \dot{x}~=~0. \tag{v0}\label{eq:v0}$$
Zero velocity $\eqref{eq:v0}$ is equivalent to a constant position $$ x~=~x_0. \tag{x}\label{eq:x}$$
The solutions to zero mechanical energy $\eqref{eq:E0}$ not already covered by $\eqref{eq:N2}$ are a constant position $\eqref{eq:x}$ with vanishing potential energy $$ V(x_0)~=~0. \tag{V0} \label{eq:V0}$$
To avoid the loophole $\eqref{eq:V0}$ let us additionally assume that the potential energy $V$ is non-vanishing everywhere $$ \forall x:~~ V(x)~\neq~0. \tag{V}\label{eq:V}$$ Then the Lagrangian $L$ is equivalent to the usual $T-V$ at the classical level.
We formulate a new type of the Lagrangian called the multiplicative one $L(\dot x,x)=F(\dot x)G(x)$, where $F$ and $G$ must be determined subject to the equation of motion. It turn out that new Lagrangian comes with a parameter $\lambda$
$L_\lambda(\dot x,x)=m\lambda^2\left(e^{-\frac{\dot x^2}{2\lambda^2}}+\frac{\dot x}{\lambda^2} \int_{0}^{\dot x}d\dot u e^{-\frac{\dot u^2}{2\lambda^2}}\right)e^{-\frac{V(x)}{m\lambda^2}}$
and under the limit $\lambda\rightarrow \infty$, the standard Lagrangian is recovered. An interesting point is that if we expand the new Lagrangian with respect to the parameter we obtain a Lagrangian hierarchy
$L_1=T-V$
$L_2=T^2/2 -2T^2V+V^2$
$L_3=\frac{1}{5}T^3(\dot x)+T^2(\dot x)V(x)+3T(\dot x)V(x)-V^3(x)$
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This means that we find a non-trivial way to generate Lagrangians which give the same equation of motion. Indeed, the physical meaning still remains to investigate further, but at least, we knew how to produce this type of Lagrangian, and it does not come out of nowhere.
In this answer we expand on sikarin yoo-kong's answer.
Define 1D Lagrangian on factorized/multiplicative form: $$ L(x,v) ~=~F(v)G(x),\qquad v~\equiv~\dot{x}. \tag{1}$$
Euler-Lagrange equation: $$\begin{align} F(v)G^{\prime}(x) ~=~&\frac{\partial L}{\partial x} ~=~\frac{d}{dt}\frac{\partial L}{\partial v} ~=~\frac{d}{dt}\left[F^{\prime}(v)G(x)\right]\cr ~=~&\dot{v}F^{\prime\prime}(v)G(x)+F^{\prime}(v)vG^{\prime}(x). \end{align}\tag{2}$$
Since we would like the Lagrangian (1) to encode Newton's 2nd law $$m\dot{v}~=~-V^{\prime}(x), \tag{3}$$ we are lead to the following separable equation $$ -\frac{G^{\prime}(x)}{V^{\prime}(x)G(x)}~=~\frac{1}{E}~=~\frac{1}{m}\frac{F^{\prime\prime}(v)}{F(v)-vF^{\prime}(v)}, \tag{4}$$ where $E$ is a separation constant with dimension of energy.
In other words, we have a 1st-order homogeneous ODE $$EG^{\prime}(x)~=~-V^{\prime}(x)G(x) \tag{5}$$ with solution $$G(x)~=~e^{-\frac{V(x)}{E}}; \tag{6} $$ and we have a 2nd-order homogeneous ODE $$\frac{E}{m}F^{\prime\prime}(v)~=~F(v)-vF^{\prime}(v), \tag{7}$$ or equivalently $$\frac{d}{dv}\left(e^{\frac{T(v)}{E}}F^{\prime}(v)\right)~=~\frac{m}{E}e^{\frac{T(v)}{E}}F(v), \qquad T(v)~\equiv~\frac{m}{2}v^2. \tag{8}$$
The 1st solution $$F_1(v)~=~v \tag{9}$$ to the ODE (7) is not interesting.
The Wronskian $$ W(v)~\equiv~F_1(v)F_2^{\prime}(v)- F_1^{\prime}(v)F_2(v)\tag{10}$$ satisfies the 1st-order homogeneous ODE $$ W^{\prime}(v)+\frac{mv}{E}W(v)~=~0\tag{11}$$ with solution $$ W(v)~=~e^{-\frac{T(v)}{E}}.\tag{12}$$
The 2nd solution to the ODE (7) is then $$\begin{align}F_2(v)~=~& -F_1(v)\int\!dv\frac{W(v)}{F_1(v)^2}\cr ~=~&-v\int\!dv\frac{W(v)}{v^2} ~\stackrel{\rm IBP}{=}~W(v)+\frac{mv}{E}\int\!dv~W(v)\cr ~=~&-\frac{T^{1/2}}{2}\int\!dT\frac{W(T)}{T^{3/2}} ~=~-\sum_{n=0}^{\infty}\frac{1}{n!(2n\!-\!1)}\left(-\frac{T(v)}{E}\right)^n\cr ~=~&1+\frac{T}{E}-\frac{T^2}{6E^2}+\frac{T^3}{30E^3}+\ldots .\end{align}\tag{13}$$
The factorized Lagrangian (1) for the 2nd solution then expands into an infinite tower/hierarchy of equivalent Lagrangians $$\begin{align} L~=~&F_2(v)G(x)\cr ~=~&\left(1+\frac{T}{E}-\frac{T^2}{6E^2}+\frac{T^3}{30E^3}+\ldots\right)\left(1-\frac{V}{E}+\frac{V^2}{2E^2}-\frac{V^3}{6E^3}+\ldots\right)\cr ~=~&1+\frac{T-V}{E}+\frac{-\frac{T^2}{3}-2TV+V^2}{2E^2}+\frac{\frac{T^3}{5}+T^2V+3TV^2-V^3}{6E^3}+\ldots. \end{align}\tag{14}$$ The standard Lagrangian appears at 1st order ${\cal O}(E^{-1})$. OP's Lagrangian appears at 2nd order ${\cal O}(E^{-2})$.
The Euler-Lagrange equation (2) for the factorized Lagrangian (1) simplifies to $$\begin{align} 0~=~&\frac{\partial L}{\partial x} -\frac{d}{dt}\frac{\partial L}{\partial v}\cr ~=~&[F(v)-vF^{\prime}(v)]G^{\prime}(x) -\dot{v}F^{\prime\prime}(v)G(x)\cr ~=~&\frac{m\dot{v}+V^{\prime}(x)}{E} [vF^{\prime}_2(v)-F_2(v)]G(x)\cr ~=~&\frac{m\dot{v}+V^{\prime}(x)}{E}W(v)G(x)\cr ~=~&\frac{m\dot{v}+V^{\prime}(x)}{E} e^{-\frac{T(v)+V(x)}{E}} .\end{align}\tag{15}$$ At the $n$th order ${\cal O}(E^{-n})$ eq. (15) is equivalent to either (i) Newton's 2nd law (3) or to (ii) zero mechanical energy $$ T(v)+V(x)~=~0. \tag{16}$$ This situation is analyzed in my other Phys.SE answer.
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