alleryIn decomposing $\dfrac{1}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}$, the constants are
A$A=1,\ B=-1$
B$A=-1,\ B=1$
C$A=1,\ B=1$
D$A=2,\ B=-1$
Answer & Solution
Correct answer: A. $A=1,\ B=-1$
1. Clear denominators: $1=A(x+2)+B(x+1)$.
2. Put $x=-1$: $1=A(1)$, so $A=1$.
3. Put $x=-2$: $1=B(-1)$, so $B=-1$.
4. Thus $A=1,\ B=-1$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.317_
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