= Solution
In the steady state, <mass conservation> reduces to $q_x=R$. The zero-flux condition at the divide gives $q=R(x-L)$. Substituting the <Darcy flux> and integrating from the absorbing river boundary yields
$$
u_b\left(\frac{h^2}{2}+\frac{\beta h^3}{6}\right)=R\left(Lx-\frac{x^2}{2}\right).
$$
Thus the nonnegative groundwater height is the unique nonnegative root of
$$
\boxed{h^2+\frac\beta3h^3=\frac R{u_b}(2Lx-x^2).}
$$
The height increases toward the divide, where $h_x=0$. The river receives all the rainfall, so its steady <volume flux per unit width> is $\boxed{Q=RL}$.
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