In 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: NCERT Class 12 Mathematics Ch 7 "Integrals", p.317_
Related questions
$\displaystyle\int_2^3 \dfrac{x\,dx}{x^2+1}$ equals$\displaystyle\int_0^{2a} f(x)\,dx$ equals $2\displaystyle\int_0^{a} f(x)\,dx$ precisely wIf $f(a+b-x)=f(x)$, then $\displaystyle\int_a^b x\,f(x)\,dx$ is equal to$\displaystyle\int \frac{dx}{e^{x}+e^{-x}}$ is equal toA rational function $\dfrac{P(x)}{Q(x)}$ is called proper when$\displaystyle\int_0^{\pi/4} \tan x\,dx$ equals$\displaystyle\int_0^{1} x e^{x^2}\,dx$ equalsThe value of $\displaystyle\int_0^{\pi/2} \log\!\left(\dfrac{4+3 in x}{4+3\cos x}\right)dx