When decomposing $\dfrac{px^2+qx+r}{(x-a)^2(x-b)}$ into partial fractions, the correct form is
A$\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{(x-b)^2}$
B$\dfrac{A}{x-a}+\dfrac{Bx+C}{(x-a)^2}+\dfrac{D}{x-b}$
C$\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}$
D$\dfrac{Ax+B}{(x-a)^2}+\dfrac{Cx+D}{x-b}$
Answer & Solution
Correct answer: C. $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}$
1. A repeated linear factor $(x-a)^2$ contributes two terms: $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$.
2. The distinct factor $(x-b)$ contributes one term $\dfrac{C}{x-b}$.
3. Option A wrongly repeats $(x-b)$ and ignores the repetition of $(x-a)$.
4. Options B and D wrongly place linear numerators over a repeated linear factor; the form is $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}$ (Table 7.2 case 4).
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.316_
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