$A$ and $B$ are constants such that
$$\frac{4x+5}{x^2+x-2}= \frac{A}{x+2} +\frac{B}{x-1}. $$Find $$\frac{B}{x+1} - \frac{A}{x-2}.$$
We have that
$$\frac{4x+5}{x^2+x-2}= \frac{4x+5}{(x+1)(x-2)}=\frac{A}{x+2} +\frac{B}{x-1}. $$Multiplying both sides by $(x+1)(x-2)$ gives
$$4x+5=A(x+1)+B(x-2)$$which we can expand to get
$$4x+5=(A+B)x+A-2B.$$By comparing coefficients, we see that $A+B=4$ and $A-2B=5$. We can solve these equations to get $A=1$ and $B=3$.
Then
$$\begin{aligned} \frac{B}{x+1} - \frac{A}{x-2}&=\frac{3}{x+1} - \frac{1}{x-2}\\
&=\frac{3(x-2)-1(x+1)}{(x+1)(x-2)}\\
&=\boxed{\frac{2x-7}{(x+1)(x-2)}}.
\end{aligned}$$