\begin{aligned}
\frac{3}{x^2+x-2}=\frac{3}{(x+2)(x-1)} &=\frac{A}{x+2}+\frac{B}{x-1} \\
&=\frac{A(x-1)+B(x+2)}{(x+2)(x-1)}
\end{aligned}
Then,
$$
\mathrm{A}(x-1)+\mathrm{B}(x+2)=3
$$
when \(x=1\)
$$
\mathrm{A}(1-1)+\mathrm{B}(1+2)=3
$$
$$
\begin{array}{l}
\mathrm{A}(0)+3 \mathrm{~B}=3 \\
3 \mathrm{~B}=3, \mathrm{~B}=3 / 3=1
\end{array}
$$
when \(x=-2\)
$$
\begin{array}{l}(-2-1)+B(-2+2)=3 \\
-3 A+B(0)=3 \\
-3 A=3, A=3 /-3=-1
\end{array}
$$
Then,
$$
\begin{aligned}
\frac{3}{x^2+x-2} &=\frac{-1}{x+2}+\frac{1}{x-1} \\
&=\frac{1}{x-1}-\frac{1}{x+2}
\end{aligned}
$$