# 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.

• Substitute $u=\dot x$? Commented Aug 20, 2022 at 13:35
• $~\dot x\ddot x=\dfrac{d}{dt}\left( \dfrac{1}{2}~\dot x^{2}\right) ~$
– Eli
Commented Aug 20, 2022 at 13:39
• @WillO I wrote in the post that I'm trying to solve it without $\frac{d}{dt}f(g(x))=f'(g(x))\cdot g'(x)$, which is what Eli suggested
– Yoav
Commented Aug 20, 2022 at 13:56
• @GedankenExperimentalist thank you for your help!
– Yoav
Commented Aug 20, 2022 at 14:00
• @MeepMeep You need to revise the difference between the product rule and the chain rule. In any case, restricting yourself to solving an integral without using one or the other of these basic tools seems pointless and arbitrary. Commented Aug 20, 2022 at 15:17

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.