Integrating a product, one factor a derivative ∫dx(t)dtx(t)2dt

If integration by parts gave you 0, its

I’m going to assume that ${\displaystyle \frac{x\left(t\right)}{dt}}$ is a typo for ${\displaystyle \frac{d\left(x\left(t\right)\right)}{dt}}$.
Look at a specific example, say with ${\displaystyle x\left(t\right)=\sin t}$. Then
${\displaystyle \int \frac{d\left(x\left(t\right)\right)}{dt}x{\left(t\right)}^{2}dt=\int \cos t{\sin }^{2}tdt}$,
a problem that you would most likely solve by making the substitution ${\displaystyle u=\sin t,du=\cos tdt}$, and integrating
${\displaystyle \int u^{2}du}$.
But you don’t have to know what x(t) is to make this substitution. If you let ${\displaystyle u=x\left(t\right)}$, then
${\displaystyle du=\frac{d\left(x\left(t\right)\right)}{dt}dt}$,
and ${\displaystyle \int \frac{d\left(x\left(t\right)\right)}{dt}x{\left(t\right)}^{2}dt=\int u^{2}du=\frac{u^3}{3}+C=\frac{x{\left(t\right)}^3}{3}+C}$

Also, since $\int \frac{dx(t)}{dt}x(t{)}^{2}dt=\int \frac{df(t)}{dt}dt$, where $f(t)=\frac{1}{3}x(t{)}^3$, we can use the Second Fundamental Theorem of Calculus to conclude that $\int \frac{dx(t)}{dt}x(t{)}^{2}dt=f(t)+C=\frac{1}{3}x(t{)}^3+C$. No substitution needed.