= Solution
Choose the convention $L_xG(x,s)=\delta(x-s)$, where $\delta$ is the <Dirac delta distribution>, together with $G(1/2,s)=G(1,s)=0$. The <Green's function> is continuous at $x=s$ and obeys
$$
\partial_xG(s^+,s)-\partial_xG(s^-,s)=1.
$$
These conditions follow by integrating its <differential equation> across $s$; continuity prevents a derivative of the <Dirac delta distribution>. If $\ell$ and $r$ are homogeneous solutions satisfying the left and right <boundary conditions>, respectively, and $W=\ell r'-\ell'r$ is their <Wronskian>, then
$$
G(x,s)=\frac{\ell(x_<)r(x_>)}{W(s)},\qquad x_<=\min(x,s),\quad x_>=\max(x,s).
$$
For $n\ne0$ take $\ell(x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$. Their <Wronskian> is $W(s)=n(2^n-1)s^{-n-1}$, and hence
$$
\boxed{G_n(x,s)=\frac{s^{n+1}(x_<^{-n}-2^n)(x_>^{-n}-1)}{n(2^n-1)},\qquad n\ne0.}
$$
For $n=0$ take $\ell(x)=\log(2x)$ and $r(x)=\log x$, giving $W(s)=\log2/s$ and
$$
\boxed{G_0(x,s)=\frac{s}{\log2}\log(2x_<)\log x_>.}
$$
Both formulas satisfy the endpoint conditions, continuity, and the unit jump directly. The unique solution of the <boundary value problem> is
$$
\boxed{u(x)=\int_{1/2}^1G_n(x,s)f(s)\,ds.}
$$
The factor $s^{n+1}$ in the first formula matters: this <Green's function> is for the original monic <differential operator>, with ordinary measure $ds$, rather than for the operator multiplied by $x^{n+1}$.
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