1.59 Determine $A, B$ and $C$ so that for all permissible values of $x$ the equality holds

$\frac{x^{2}+5}{x^{3}-3 x+2}=\frac{A}{x+2}+\frac{B}{(x-1)^{2}}+\frac{C}{x-1}$.
1.59 Let's bring the right-hand side of the equation to a common denominator:

\(\frac{x^{2}+5}{x^{3}-3 x+2} \equiv \frac{A(x-1)^{2}+B(x+2)+C(x+2)(x-1)}{x^{3}-3 x+2}\).

If two fractions with the same denominator are equal, then their numerators are also identically equal, i.e.,

\[
x^{2}+5 \equiv A(x-1)^{2}+B(x+2)+C(x+2)(x-1)
\]

Since this equality is an identity, it holds for any \(x\). Therefore, if \(x=1\), then \(3 B=6\), i.e., \(B=2\); if \(x=-2\), then \(9 A=9\), i.e., \(A=1\); if \(x=0\), then \(A+2 B-2 C=5\), from which \(C=0\).

Omeem: \quad A=1, B=2, C=0\).