I3.1 Given that $\frac{2 x-3}{x^{2}-x}=\frac{A}{x-1}+\frac{B}{x}$, where $A$ and $B$ are constants. If $S=A^{2}+B^{2}$, find the value of $S$
$\begin{array}{l}\frac{2 x-3}{x^{2}-x}=\frac{A x+B(x-1)}{(x-1) x} \\ A+B=2,-B=-3 \\ A=-1, B=3 \\ S=(-1)^{2}+3^{2}=10\end{array}$