Question

If $\frac{3x^3+5x-4}{(x^2-1)(x^2-4)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x+2}+\frac{D}{x-2}$, then $A+B+C+D$ is

• A. 4
• B. 3
• C. 0
• D. 2

Answer 1

The correct option is B 3

$\frac{3x^3+5x-4}{(x^2-1)(x^2-4)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x+2}+\frac{D}{x-2}$
$\frac{3x^3+5x-4}{(x^2-1)(x^2-4)}=\frac{3x^3+5x-4}{(x-1)(x+1)(x+2)(x-2)}$
Resolving into partial fractions,
$\frac{3x^3+5x-4}{(x-1)(x+1)(x+2)(x-2)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x+2}+\frac{D}{x-2}$

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

When $x=1\Rightarrow 3+5-4=0+B(2\cdot 3\cdot -1)+0\Rightarrow B=-\frac{2}{3}$

When $x=-1\Rightarrow -12=A\left(6\right)\Rightarrow A=-2$

When $x=2\Rightarrow 30=D\left(12\right)\Rightarrow D=\frac{5}{2}$

When $x=-2\Rightarrow -38=C(-12)\Rightarrow C=\frac{19}{6}$

So, $A+B+C+D=\frac{-2}{3}-2+\frac{5}{2}+\frac{19}{6}=3$.