 Solution

On each interval of width $h=1/N$, Taylor expansion about its midpoint shows that the local midpoint-rule error is $O(h^3)$. Summing over $N$ intervals gives the composite midpoint rule error

To remove the endpoint singularity, set $x=s^2$. Thenwhose transformed integrand is analytic. Applying the composite midpoint rule in $s$ with $s_n=(n+\frac{1}{2})/N$ givesThis has the required form whenThe quadratic substitution for a square-root endpoint singularity therefore restores second-order convergence.