Question

Resolve into partial fraction $\frac{(2x-5)}{(x+3)(x+1{)}^2}$

• A. $\frac{11}{4(x+3)}-\frac{7}{2(x+1{)}^2}+\frac{11}{4(x+1)}$
• B. $-\frac{11}{4(x+3)}-\frac{7}{2(x+1{)}^2}+\frac{11}{4(x+1)}$
• C. $-\frac{11}{4(x+3)}+\frac{7}{2(x+1{)}^2}+\frac{11}{4(x+1)}$
• D. $\frac{11}{4(x+3)}+\frac{7}{2(x+1{)}^2}+\frac{11}{4(x+1)}$

Answer 1

Answer: (B)

$-\frac{11}{4(x+3)}-\frac{7}{2(x+1{)}^2}+\frac{11}{4(x+1)}$

Let $\frac{(2x-5)}{(x+3){(x+1)}^2}=\frac{A}{x+3}+\frac{B}{x+1}+\frac{C}{{(x+1)}^2}$
$\Rightarrow 2x-5=A{(x+1)}^2+B(x+1)(x+3)+C(x+3)$

$\Rightarrow 2x-5=A(x^2+2x+1)+B(x^2+4x+3)+C(x+3)$

On comparing coefficients

$0=A+B,2=2A+4B+C,-5=A+3B+3C$

$\Rightarrow A=-\frac{11}{4},B=\frac{11}{4},C=-\frac{7}{2}$

Hence, $\frac{(2x-5)}{(x+3){(x+1)}^2}=-\frac{11}{4(x+3)}+\frac{11}{4(x+1)}-\frac{7}{2{(x+1)}^2}$

Hence, option 'B' is correct.