Trying to solve second equation of motion without substituting $v=(at+v_i)$ [closed]

Question

The derivation of $s = s_0 + v_0t + ½at^2$ starts with $v = dx/dt$, which is rewritten as $v dt = dx$ and then $v$ is replaced with $at+v_i$. Then it is integrated.

1. Why is this substitution done, instead of integrating $v dt = dx$ directly? Is it to replace $v$ (not constant) with $a$ and $v_i$ (constants) and therefore make the integration easier?
2. If $v = dx/dt$ were integrated instead, would you have to use integration by parts (because $v(t)$ is not constant wrt to time)? When I try this, I cannot get the final form of the equation ($x=x_0 +v_0t+1/2at^2$) that one gets when using the $at+v_i$ substitution. (Could be math error, but I get a $-1/2at^2$ term)

The reason I ask is because making the substitution is not intuitive to me (unless it's for #1 as stated above). Could someone explain the reason for the substitution? Could someone explain why $v = dx/dt$ can't be integrated directly? Or if it can, can you show how it's done please.

Answer 1

For clarification, the case you are considering is for constant acceleration; that is, a is constant. v(t) is not constant, so to evaluate $\int_{}^{}v(t) \thinspace dt$ you use $v(t) = v_i + at$ as you state.

When you integrate by parts $\int_{s_0}^{s} dx = s - s_0 =$ $\int_{0}^{t}v \enspace dt = (vt)|{_i ^f} - \int_{v_0}^{v_f}t \enspace dv$ you do obtain $-1/2 \enspace a t^2$ for the second term since dv = a dt, but you have to evaluate the first term, vt, between the initial, i, and final, f, states; initially, vt = 0 and finally vt = $v_ft = (v_0 + at)t $ and adding this to the second term you have $(v_0 + at)t -1/2 \enspace a t^2 = v_0t + 1/2 \enspace a t^2$