I have the following demonstration
$\langle x|x'\rangle = \langle x|(\int{|p\rangle \langle p|dp})|x'\rangle = \int{\langle x|p\rangle\overline{\langle x'|p\rangle}dx}=\int{e_{p}(x)\overline{e_{p}(x')}dx}=\delta(x-x')=\langle x|x'\rangle=\langle x|\cdot 1\cdot|x'\rangle$
I don't understand how in the second step we go from having an integral on dp to an integral on dx, if someone could explain I would appreciate it.