First let me quickly explain what angular velocity is.
Consider two right-handed, orthonormal coordinate systems with coincident origins. We will consider one of these coordinate systems to be the "world" (some kind of reference) and the other to be the "body" (the thing who's orientation we are concerned with).
A point $p$ can be expressed in either world coordinates or body coordinates, related by a rotation matrix.
$$p_w = R\ p_b$$
In this case, I have defined $R$ to convert body coordinates to world coordinates, but of course its inverse will perform the reverse transformation.
When we say that the body is "rotating" relative to the world, what we mean is that $R = R(t)$ is really a function of a scalar called time. So now lets take derivatives and employ the chain rule. Let $\dot{x} := \frac{dx}{dt}$.
$$\dot{p}_w = R\ \dot{p}_b + \dot{R}\ p_b$$
What can be said about $\dot{R}$? This is where we have to make use of our choice of consistently-handed orthonormal coordinate systems, which is practically always the case. Under this assumption, $R$ is an orthonormal matrix, meaning $R^{-1}=R^T$ or,
$$R^TR = I$$
Taking the derivative and employing the chain rule again,
$$R^T\dot{R} + \dot{R}^TR = 0$$
Which implies,
$$R^T\dot{R} = -\dot{R}^TR$$
The derivative and transpose obviously commute, so we realize that the right-hand-side is equal to the negative-transpose of the left-hand-side. This means that the matrix $R^T\dot{R}$ is skew-symmetric. Let's name it,
$$\Omega := R^T\dot{R}$$
This is called the angular velocity matrix (in body coordinates), and being as it is a skew-symmetric 3x3 matrix, it only has 3 unique elements, so we can extract them and call it the angular velocity vector,
$$\omega := \begin{bmatrix} \Omega_{2,1} \\ \Omega_{3,1} \\ \Omega_{3,2} \end{bmatrix}$$
for which the cross-product operation is equivalent to matrix multiplication,
$$\omega \times v \equiv \Omega v,\ \ \forall v \in \mathbb{R}^3$$
Coming back to our original problem, we can substitute in an $RR^T=I$ to see,
\begin{align*}
\dot{p}_w &= R\ \dot{p}_b + RR^T\dot{R}\ p_b\\
&= R\ \dot{p}_b + R\ \Omega\ p_b\\
&= R(\dot{p}_b + \omega \times p_b)\\
\end{align*}
which is sometimes called the transport theorem, but is more just the very recognizable equation for relating velocities in different reference frames. You must "augment" $\dot{p}_b$ with its rotational velocity $\omega \times p_b$ before you can convert it to world coordinates. This is my "proof" to you that the $\omega$ I have derived here is the same $\omega$ you are familiar with.
Now then, we are ready to talk about the problems of "integrating" angular velocity. Considering,
$$\Omega := R^T\dot{R}$$
we are reminded that we don't really care about the integral of $\Omega$. We only care about the integral of $\dot{R}$. Worded another way, the orientation, $R$, of the body evolves as,
$$\dot{R} = R \Omega$$
Since $\Omega$ itself is also a function of time, this linear differential equation has a solution in terms of a (non-commutative) product integral,
$$R(t) = R(t_0) \prod_{t_0}^t e^{\Omega(\tau) d\tau}$$
Notice that the matrix exponential of skew-symmetric matrices is always orthogonal, and the product of orthogonal matrices is always orthogonal (plenty of proofs out there). Thus the above equation is elegantly consistent.
Anyway, the key point here is that the angular velocity integral (in the exponent) is meaningless on its own, and as many people have pointed out is not on its own equal to change in orientation. Instead, it is related to change in orientation through that above equation.
Quaternions can come into play if you change $\Omega$ to $\omega$ in the exponent and consider the quaternion version of Euler's identity.
If your two coordinate systems do not have coincident origins, or rather are translating with respect to each other, the very first equation I wrote will have an "affine" offset, which you can deal with in many ways, but I'd leave that to you. It doesn't affect the evolution of $R$.
The angle-axis representation that you mention in your question can be introduced as follows. If we consider $\Omega$ to be constant over some (likely short) timestep, $\Delta t$, then,
$$R(t_0+\Delta t) = R(t_0)e^{\Omega \Delta t}$$
The orthogonal marix $e^{\Omega \Delta t}$ can be considered a "rotation caused by $\Omega$" and it will have a central axis aligned with $\omega$ and an angle of rotation equal to $||\omega||\Delta t$. This is the axis-angle interpretation of $\omega$ your book refers to. It can be used to "step forward" $R$ by $\Delta t$ seconds, where $\Delta t$ is the period overwhich $\Omega$ is constant.
Final thing to mention: $e^{\Omega \Delta t}$ is, in practice, computed using Rodriguez's formula instead of taking an actual matrix exponential, but both do work.
