Here is another way to look at where $L$. For a dissipative system it is more natural to write the Lagrange just in terms of the kinetic energy $T$ instead of $L$:
$$\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} = F_i = \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial q^i}$$
This is the equation that the Beers actually derives.
It is the general form. The conservative case is obtained by setting $\mathbf{F} = -\nabla V$.
\begin{align} \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} &= - \frac{\partial V}{\partial q_i}, \text{ since } \frac{\partial V}{\partial q_i} = \nabla V \cdot \frac{\partial \mathbf{r}}{\partial q^i} \\ \therefore \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial (T - V) }{\partial q_i} &= 0 \\ \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \end{align}

Now $\frac{\partial V}{\partial \dot{q}_i} = 0$ since by definition $V$ is a function of only the $q_i$s and thus independent of $\dot{q}_i$, so
\begin{align} \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial V}{\partial \dot{q}_i} \right) - \frac{\partial L }{\partial q_i} &= 0 \\ \frac{d}{dt} \frac{\partial (T - V)}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \\ \end{align}

Seen this way $L$ looks like a bit of a hack.
It is really an artifact of the system being conservative and even then replacing $T$ by $L$ in the first term of the equation is kind of redundant.