= Solution
For $h=1/N$, the <composite trapezoidal rule> is
$$
T_N(f)=\frac1N\left(\frac{f(0)}2+\sum_{m=1}^{N-1}f(m/N)+\frac{f(1)}2\right).
$$
With $f(x)=1/(4-x^2)$ this becomes
$$
T_N=\frac{7}{24N}+\sum_{m=1}^{N-1}\frac{N}{4N^2-m^2}.
$$
Write the requested right <Riemann sum> as $S_N$. The <trapezoidal endpoint correction for a right Riemann sum> gives
$$
S_N-T_N=\frac{f(1)-f(0)}{2N}=\frac1{24N}.
$$
The integrand is continuous on $[0,1]$, so the trapezoidal values converge to its definite integral; the endpoint correction tends to zero. Therefore
$$
\boxed{\lim_{N\to\infty}\sum_{m=1}^N\frac{N}{4N^2-m^2}=\frac14\log3.}
$$
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