Question

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

• A. $-\frac{11}{4(x+3)}+\frac{11}{4(x+1)}-\frac{7}{2(x+1{)}^2}$
• B. $\frac{11}{4(x+3)}-\frac{11}{4(x+1)}-\frac{7}{2(x+1{)}^2}$
• C. $-\frac{11}{4(x+3)}+\frac{11}{4(x+1)}-\frac{5}{2(x+1{)}^2}$
• D. $-\frac{11}{4(x+3)}+\frac{11}{4(x+1)}+\frac{5}{2(x+1{)}^2}$

Answer 1

The correct option is A $-\frac{11}{4(x+3)}+\frac{11}{4(x+1)}-\frac{7}{2(x+1{)}^2}$

Let $\frac{(2x-5)}{(x+3)(x+1{)}^2}=\frac{A}{x+3}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$ ...(i)
$⟹\frac{(2x-5)}{(x+3)(x+1{)}^2}=\frac{A(x+1{)}^2+B(x+3\left)\right(x+1)+C(x+3)}{(x+3)(x+1{)}^2}$
$⟹2x-5=A\left(\right(x+1{)}^2+B(x+3)(x+1)+C(x+3)$ ...(ii)

Putting $x+1=0$ or $x=-1$ in (ii), we obtain $-2-5=0+0+C(-1+3)$

$\therefore C=-\frac{7}{2}$

Putting $x+3=0$ or $x=-3$ in (ii), we obtain $-6-5=A(-3+1{)}^2+0+0$

$\therefore A=-\frac{11}{4}$

Equating the coefficient of $x^2$ on both sides of (ii), then $0=A+B$

$\therefore B=-A=\frac{11}{4}$

Substituting the value of a, B and C in (i), then

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

which are the required partial fractions.

Hence, option 'A' is correct.