We define $$I[f] := \int_0^1 f(x) \text dx$$
Let $\alpha \in (-1,1)\backslash\{0\}$ and $f_\alpha(x) := x^\alpha$. With $\textbf x := (x_j)^N_{j=1}, 0 \leq x_1 < \dots < x_N \leq 1$ we define the composite midpoint method: $$M_{\textbf x}[f] := \sum_{j=1}^{N - 1}(x_{j+1} - x_j) \cdot f\left(\frac{x_{j+1} + x_j}{2}\right)$$ I now want to find points $x_j^N \in (0,1]$, such that we obtain with $\textbf x^N := (x_j^N)_{j=1}^N$ for every $N \in \mathbb N$ the following inequality: $$|I[f_\alpha] - M_{\textbf x^N}[f_\alpha]| \leq C_\alpha N^{-2}$$ where $C_\alpha$ is a constant independent of $N$.
I tried setting $x_j^N := \frac 1 N$, but that did not yield anything promising and now I am out of ideas, so I appreciate any hints!