= Solution
Write $r=e^x$, $\epsilon=\hbar/\sqrt{2m}$, $L=l+1/2$, and $\lambda=-|\lambda|$. Since $V(r)=-V_0/r$, the transformed equation has
$$
q(x)=|\lambda|e^{2x}-V_0e^x+\epsilon^2L^2.
$$
The two turning radii $a<b$ are the positive roots of
$$
|\lambda|r^2-V_0r+\epsilon^2L^2
=|\lambda|(r-a)(r-b),
$$
so
$$
a+b=\frac{V_0}{|\lambda|},
\qquad
ab=\frac{\epsilon^2L^2}{|\lambda|}.
$$
Inside the allowed interval,
$$
|q|=|\lambda|(r-a)(b-r),
\qquad dx=\frac{dr}{r}.
$$
The <semiclassical action integral> is therefore
$$
\begin{aligned}
\frac1\epsilon\int_{x_a}^{x_b}|q(x)|^{1/2}\,dx
&=\frac{\sqrt{|\lambda|}}{\epsilon}
\int_a^b\frac{\sqrt{(r-a)(b-r)}}r\,dr\\
&=\frac{\pi\sqrt{|\lambda|}}{2\epsilon}(\sqrt b-\sqrt a)^2\\
&=\pi\left(\frac{V_0}{2\epsilon\sqrt{|\lambda|}}-L\right).
\end{aligned}
$$
Equating this with $(n+1/2)\pi$ from part (a), and using $L=l+1/2$, gives
$$
\frac{V_0}{2\epsilon\sqrt{|\lambda|}}=n+l+1.
$$
Since $\epsilon^2=\hbar^2/(2m)$,
$$
|\lambda|=\frac{V_0^2}{4\epsilon^2(n+l+1)^2}
=\frac{mV_0^2}{2\hbar^2(n+l+1)^2}.
$$
Thus the <Coulomb WKB quantization with the Langer correction> yields
$$
\boxed{E=\lambda=-\frac{mV_0^2}{2\hbar^2(n+l+1)^2}.}
$$
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