The Theorem 1.5 and 1.6 you quote give the answer.

More precisely, for $SO$, in the range $d<6$, the only polynomial generators
are $p_1$ which has degree 4, $\delta(w_2)$ with degree 3 and $\delta(w_4)$ with degree 5. The only relations are $2\delta(w_{2i})=0$, which gives $$H^d(BSO_{\infty};\mathbb{Z})\cong 0,\quad (d=1,2),$$
$$H^d(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2 ,\quad (d=3,5), $$
$$H^4(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}.$$

In the case of $BO$, there are more generators $\delta(w_1)$ and
$\delta(w_1w_2)$ in degrees 2 and 4. Thus in degrees 4 and 5 we also have products $\delta(w_1)^2$ and $\delta(w_1)\delta(w_2)$.

All of these lead to
$$H^1(BO_{\infty};\mathbb{Z})\cong 0,$$
$$H^i(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2, \quad (i=2,3)$$
$$H^4(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}\oplus (\mathbb{Z}/2)^2,$$
$$H^5(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

ringis simpler in the infinite case. $\endgroup$ – Mark Grant May 5 '14 at 11:54