Question

If $\frac{x^4+24x^2+28}{(x^2+1{)}^3}$
$=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1{)}^2}+\frac{Ex+F}{(x^2+1{)}^3}$ then $A=$

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (C)

$0$

let us solve through substitution

$x^2+1=k\Rightarrow x^2=k-1$

$\frac{(k-1{)}^2+24(k-1{)}^2+28}{k^3}=\frac{25k^2}{k^3}+$(other terms)

solving we get

$=\frac{25}{k}+$(other terms)

$=\frac{25}{(x^2+1)}+$(other terms)

$[k=x^2+1]$

So, $A=0;B=25$

$A=0$