Question

If $\frac{x+4}{(x^2-4)(x+1)}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{x+1}$, then $A+B+C=$

• A. 2
• B. 0
• C. -1
• D. 1

Answer 1

The correct option is B 0

Given, $\frac{x+4}{(x^2-4)(x+1)}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{x+1}$

LHS $=\frac{x+4}{(x^2-4)(x+1)}$
RHS $=\frac{A(x+2)(x+1)+B(x-2)(x+1)+C(x-2)(x+2)}{(x^2-4)(x+1)}$
LHS $=$ RHS
For any value of $x$, they must be equal.
So, finding the values of $A,B,C$
LHS $=$ RHS
$(x+4)=A(x+2)(x+1)+B(x-2)(x+1)+C(x-2)(x+2)$
Let $x=2$
Substituting in the equation, we get

$A=\frac{1}{2}$

Let $x=-2$, we get $B=\frac{1}{2}$
and let $x=-1$, we get $C=-1$
So, $A+B+C=0$