= Solution
Let $H_0|r\rangle=E_r|r\rangle$. In the <interaction picture>,
$$
\delta H_I(t)
=e^{iH_0t/\hbar}\delta H(t)e^{-iH_0t/\hbar},
$$
and the state obeys
$$
i\hbar\frac d{dt}|\psi_I(t)\rangle
=\delta H_I(t)|\psi_I(t)\rangle.
$$
The first term of the <Dyson series>, or equivalently first-order <time-dependent perturbation theory>, gives
$$
|\psi_I(t)\rangle
=\left[
1-\frac i\hbar\int_0^t\delta H_I(t')\,dt'
\right]|\psi(0)\rangle
+O(\delta H^2).
$$
Thus, if $|\psi(0)\rangle=\sum_ra_r|r\rangle$,
$$
\boxed{
|\psi_I(t)\rangle
=\sum_s\left[
a_s-\frac i\hbar\sum_ra_r
\int_0^t
e^{i(E_s-E_r)t'/\hbar}
\langle s|\delta H(t')|r\rangle\,dt'
\right]|s\rangle
+O(\delta H^2)
}.
$$
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