We have $$ v(t') =v(0)+ \int_0^{t'} a(t_1) dt_1=v(0)+\int_0^{t'}dt_1 F(t_1)/m. $$ This follows from the fundamental theorem of calculus
$$ \frac{dv(t')}{dt'}= \frac{d}{dt'}\int_0^{t'} a(t_1) dt_1= a(t'), $$ and the initial condition $v(t=0)=v(0)$.

Next $$ x(t)= \int_0^t v(t') dt'. $$ Just put them together.

We have $$ v(t') =v(0)+ \int_0^{t'} a(t_1) dt_1=v(0)+\int_0^{t'}dt_1 F(t_1)/m. $$ This follows from the fundamental theorem of calculus
$$ \frac{dv(t')}{dt'}= \frac{d}{dt'}\int_0^{t'} a(t_1) dt_1= a(t'), $$ and the initial condition $v(t=0)=v(0)$. This inner integral goes only from $t=0$ to $t=t'$ becuase that is when we need $v(t)$ for the next integral: $$ x(t)= \int_0^t v(t') dt'. $$ Now just put them together to get $$ x(t) = \int_0^t v(t') dt'= \int_0^t \left(v(0) +\int_0^{t'}dt_1 F(t_1)/m\right ) dt'\\ =v(0)t+ \int_0^t \left(\int_0^{t'}dt_1 F(t_1)/m\right ) dt' $$

We have that $$ v(t') =\int_0^{t'} a(t_1) dt_1=\int_0^{t'}dt_1 F(t_1)/m $$ and $$ x(t)= \int_0^t v(t') dt'. $$ Just put them together.

We have $$ v(t') =v(0)+ \int_0^{t'} a(t_1) dt_1=v(0)+\int_0^{t'}dt_1 F(t_1)/m. $$ This follows from the fundamental theorem of calculus
$$ \frac{dv(t')}{dt'}= \frac{d}{dt'}\int_0^{t'} a(t_1) dt_1= a(t'), $$ and the initial condition $v(t=0)=v(0)$.

Next $$ x(t)= \int_0^t v(t') dt'. $$ Just put them together.