How does one integrate $m\,\dot x\, \ddot x$? How does one integrate $m\,\dot x\, \ddot x$ ?
I'm trying to integrate $m\dot x \ddot x$ without using the product rule $\frac{d}{dt}f(g(x))=f'(g(x))\cdot g'(x)$.
I've tried partial integration, but after differentiating again to check myself, I saw that the solution was wrong.
The only method that worked was substitution, but I doubt it was ok to integrate like that:
$$ 
\int m\,\dot x\, \ddot x \,dt \quad; u = \dot x \ddot x, du= (\ddot x^2 + \dot x \dddot x) dt = \ddot x^2 dt\\
\frac{1}{\ddot x^2}\int m\,u\, du = \frac{1}{2\ddot x^2}m(u)^2= \frac{1}{2\ddot x^2} m(\dot x \, \ddot x )^2 = \frac{1}{2}m \dot x^2 
$$
Again I doubt this integration is valid and I think that by chance I got the right solution, mainly because when in cases in which m is a function of time (for instance $m(t)=\gamma t + m_0$) and using the same technique, I'm getting a wrong solution.
 A: If $m$, $\dot{x}$ and $\ddot{x}$ are arbitrary functions of time, then we have the product of three functions of time,
$$
\dot{(uvw)}=\dot{u}vw+u\dot{v}w+uv\dot{w}
$$
and it would follow that the integration by parts for such a product is,
$$\int \dot{u}vw\,\mathrm{d}t=uvw-\int u\dot{v}w\,\mathrm{d}t-\int uv\dot{w}\,\mathrm{d}t\tag{1}$$
I think we would want to let $\dot{u}=\ddot{x}\Rightarrow u=\dot{x}$, $v=\dot{x}$ and $w=m$, resulting in,
$$\int \ddot{x}\dot{x}m\,\mathrm{d}t=\dot{x}\dot{x}m-\int \dot{x}\ddot{x}m\,\mathrm{d}t-\int \dot{x}\dot{x}\dot{m}\,\mathrm{d}t\tag{2}$$
The middle term of (2) is equal to the LHS, so we end up with,
$$\int \ddot{x}\dot{x}m\,\mathrm{d}t=\frac{1}{2}m\dot{x}^2-\frac{1}{2}\int \dot{x}^2\dot{m}\,\mathrm{d}t\tag{3}$$
This cannot quite be solved without knowing the functional form of $\dot{x}$ and $m(t)$ (though clearly with the assumed $m(t)=\gamma t+m_0$, $\dot{m}=\gamma$), but at least it should be clear that if $m(t)=\text{const}$ the right most term of (3) drops out and you're left with the $m\dot{x}^2/2$ you should expect and could compute otherwise.
