Integral Derivation of kinematics formula

Question

The following is part of a derivation of the formula $v=v_0 + at$: $$dv=a\ dt$$ $$\int dv=a \int dt$$ $$ v=at+C$$ Since, a C appears on the right side on integrating $dt$ why doesn't it appear on integrating $dv$

Answer 1

There actually is a constant that appears when integrating $dv$, it just gets "absorbed" into $C$ in the final expression:

\begin{align*} dv &= a\text{ }dt \\ \int dv &= \int a\text{ }dt \\ v + c_1 &= at + c_2 \\ v &= at + (c_2 - c_1) \\ \end{align*}

Since $c_2$ and $c_1$ are just numbers, we can call the difference $c_2 - c_1$ as $C$ to get: $$ v = at + C $$

Answer 2

$$\int dv=a\int dt\implies v+c_0=at+c_1\implies v=at+c_1-c_0=at+c_2.$$