Why does integral of a derivative have a constant term tacked onto the result? $\int \Bigg(\frac{d}{dt}\bigg(\vec{A} \times \frac{d\vec{A}}{dt}\bigg)\Bigg)dt$
$\frac{d}{dt}$ and $dt$ cancels and the integral vanishes:
$=\bigg(\vec{A} \times \frac{d\vec{A}}{dt}\bigg) + C$
My question is this: if the integral vanishes by cancellation, then why do I have, do I still need to add a constant C to the result?
I was wondering if somebody could write out the steps that everybody takes for granted to explain how the constant gets there in this case...
 A: Try taking the derivative again. The derivative of the constant is zero, so adding the constant gives you the same derivative. It turns out the converse is true: if two functions have the same derivative, then they differ by a constant. In order to get all antiderivatives, you have to include the constant. 
A: Its an integration rule:
$\int f'(t)~dt = f(t) + C$
Or:
$\int \frac{df(t)}{dt}~dt = f(t) + C$
A: Your wording is far from being formal. The operator $d/dt$ should not be interpreted as a fraction, rather as a differential operator, a derivation, or a vector field depending on the context. On the other hand, the differential $dt$ is its "dual", so it does not make much sense to talk about "cancellation" of $d/dt$ and $dt$. I guess this is not the place to go into the details of the topic, but if you want to know more I will add some reference. What is more, the integral does not "vanish", derivative and integral are operators inverse of each other.
Aside from this, I assume you know basic calculus rules (which is probably why your wording is imprecise). By definition the integral of a function $f$ is another function $F$ such $F'(t) = f(t)$ for all $t$. Assuming all of this makes sense, when you find one such integral, say $F_1$, then $F_1+c$ (for $c$ constant) has the same derivative: since the derivative of a constant is zero, then
$$\frac{d}{dt}(F_1+c)(t) = \frac{d}{dt}(F_1(t)+c) = \frac{d}{dt}F_1(t)=f(t).$$
So when you find a primitive $F$ of $f$ you automatically find infinitely many others, which you can write in one line as $F+c$.
