Question

If $\frac{x^2+5}{\left(x^2+1\right)\left(x-2\right)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1},$ then $A+B+C=$

• A. -1
• B. $\frac{2}{5}$
• C. $\frac{-3}{5}$
• D. 0

Answer 1

Answer: (C)

We have,
$\frac{x^2+5}{\left(x^2+1\right)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$
$\Rightarrow x^2+5=A\left(x^2+1\right)+(Bx+C)(x-2)$
$\Rightarrow x^2+5=A{x}^2+A+B{x}^2-2Bx+Cx-2C$
Equating the coefficient of $x^2,x$ and constant
terms, we get
$41=A+B,0=-2B+C,5=A-2C$
Solving these equations, we get
$A=\frac{9}{5},B=-\frac{4}{5},C=-\frac{8}{5}$
$\therefore A+B+C=\frac{9-4-8}{5}$
$=\frac{-3}{5}$