\begin{aligned}
\frac{7 x+5}{(x+2)(x-1)} &=\frac{\mathrm{A}}{(x+2)}+\frac{\mathrm{B}}{(x-1)} \\
&=\frac{\mathrm{A}(x-1)+\mathrm{B}(x+2)}{(x+2)(x+1)}
\end{aligned}
Then,
\(7 x+5=\mathrm{A}(x-1)+\mathrm{B}(x+2)\)
since it is only B we need, a value of \(x\) that can cause the term in A to be eliminated must be used, this is \(x=1\)
when \(x=1\)
\(7(1)+5=\mathrm{A}(1-1)+\mathrm{B}(1+2)\)
\(7+5=\mathrm{A}(0)+3 \mathrm{~B}\)
\(12=0+3 B\)
\(3 B=12, B=12 / 3=4\)