Solution
= Solution
With <step size> $h=1/N$ and nodes $x_j=j/N$, the <composite trapezoidal rule> gives
$$
\boxed{T_N=\frac1N\left[\frac12+
\sum_{j=1}^{N-1}\frac{1}{1+(j/N)^2}+\frac14\right]
=\frac{3}{4N}+\sum_{j=1}^{N-1}\frac{N}{N^2+j^2},\qquad I\approx T_N.}
$$
The endpoint weights are half the interior weights, and the endpoint values are $1$ and $1/2$.