Simple indefinite integral of a vector function I am having trouble with this simple integration. I am not sure of the process or steps to follow to solve this type of problem:
If $\mathbf{V}(t)$ is a vector function of $t$, find the indefinite integral:
$$\int \Big( \mathbf{V} \times \frac{d^2\mathbf{V}}{dt^2}\Big)\hspace{1mm}dt$$
My intuition is to use integration by parts, but I'm not sure how to do this with a cross product. I am currently learning only basic triple products, and this is listed as a "challenge problem". How does one integrate a cross product?

I am doubtful that this problem can be solved by integrating by parts. Even if possible, it would be tedious since you have to separate in terms of each directional components.
Step 1:
Remember what differentiation of a cross product looks like:
$\frac{d (V \times U)}{dt} = \frac{dV}{dt} \times U + V \times \frac{dU}{dt}$. Therefore, you can reasonably suspect that $V \times \frac{d^2V}{dt^2}$ is a differential of two vectors.
Step 2:
Take a guess. Since $\frac{d (V \times U)}{dt}$ has a term $V \times \frac{dU}{dt}$, it's reasonable to think that $\frac{dU}{dt} = \frac{d^2V}{dt^2}$, hence, $U =\frac{dV}{dt}$.
Step 3: Try it out!
$\frac{d (V \times \frac{dV}{dt})}{dt} = \frac{dV}{dt} \times \frac{dV}{dt} + V \times \frac{d}{dt}(\frac{dV}{dt}) = \left|\frac{dV}{dt}\right|^2\sin(0) + V \times \frac{d}{dt}(\frac{dV}{dt}) = 0 + V\times\frac{d^2V}{dV^2}$.
Step 4: Conclude.
Therefore, $\int \Big( \mathbf{V} \times \frac{d^2\mathbf{V}}{dt^2}\Big)\hspace{1mm}dt =  V\times\frac{dV}{dt} + C$.
