Question

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

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

Answer 1

Answer: (C)

$\frac{-3}{5}$

If $\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A}{(x-2)}+\frac{Bx+C}{(x^2+1)}$

then $A+B+C=$

On Solving this ,we get

$\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A(x^2+1)+(Bx+C)(x-2)}{(x-2)(x^2+1)}$

$\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A(x^2+1)+B{x}^2-2Bx+Cx-2c}{(x-2)(x^2+1)}$

$\frac{x^2+5}{(x^2+1)(x-2)}=\frac{x^2(A+B)+x(C-2B)+A-2c}{(x-2)(x^2+1)}$

$x^2+5=x^2(A+B)+x(C-2B)+A-2c$

On Comparision this , we get

$A+B=1,A-2c=5,C-2B=0$

On Solving this we get

$A=\frac{9}{5},B=\frac{-4}{5},C=\frac{-8}{5}$

Hence , Values are

$A=\frac{9}{5},B=\frac{-4}{5},C=A=\frac{-8}{5}$

Therefore ,

$A+B+C=\frac{9}{5}-\frac{4}{5}-\frac{8}{5}$

$A+B+C=-\frac{3}{5}$