If $\frac{x^4+24x^2+28}{{\left(x^2+1\right)}^3}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{{\left(x^2+1\right)}^2}+\frac{Ex+F}{{\left(x^2+1\right)}^3}$ then the value of $A+B+C+D+E+F=$

Given $\frac{x^4+24x^2+28}{{\left(x^2+1\right)}^3}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{{\left(x^2+1\right)}^2}+\frac{Ex+F}{{\left(x^2+1\right)}^3}$

i.e. $x^4+24x^2+28=\left(Ax+B\right){\left(x^2+1\right)}^2+\left(Cx+D\right)\left(x^2+1\right)+Ex+F$

$x^4+24x^2+28=A{x}^5+2A{x}^3+Ax+B{x}^4+2B{x}^2+B+C{x}^3+D{x}^2+Cx+D+Ex+F$

$x^4+24x^2+28=A{x}^5+B{x}^4+\left(2A+C\right)x^3+\left(2B+D\right)x^2+\left(A+C+E\right)x+B+D+F$

i.e. $A=0, B=1, 2A+C=0, 2B+D=24, A+C+E=0, B+D+F=28$

$\Rightarrow A=0, B=1, C=0, D=22, E=0, F=5$

Hence, $A+B+C+D+E+F=28$