Question

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

• A. $2$
• B. $\frac{7}{4}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\frac{7}{4}$

$\frac{x^2+1}{(x^2+4)(x-2)}=\frac{Ax+B}{x^2+4}+\frac{C}{x-2}$
$\Rightarrow x^2+1=(Ax+B)(x-2)+C(x^2+4)$
Put $x=2\Rightarrow C=\frac{5}{8}$
$x=0\Rightarrow B=\frac{3}{4}$
$x=1\Rightarrow A=\frac{3}{8}$
$\Rightarrow A+B+C=\frac{7}{4}$