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If the partial fractions decomposition of $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}$ is $\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2}+\frac{C}{\left(x^2+1\right)^3}$ then $B-2 A+C=$
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25
Given partial fraction of
$\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2}+\frac{C}{\left(x^2+1\right)^3}$
$\therefore x^4+24 x^2+28=A\left(x^2+1\right)^2+B\left(x^2+1\right)+C$
Now compare the coefficient of $x^4, x^2$ constant term
$\Rightarrow A=1,24=2 A+B, 28=A+B+C$
$\Rightarrow \quad B=22, \quad C=28-1-22 \Rightarrow C=5$
Now $B-2 A+C=22-2+5=25$
$\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2}+\frac{C}{\left(x^2+1\right)^3}$
$\therefore x^4+24 x^2+28=A\left(x^2+1\right)^2+B\left(x^2+1\right)+C$
Now compare the coefficient of $x^4, x^2$ constant term
$\Rightarrow A=1,24=2 A+B, 28=A+B+C$
$\Rightarrow \quad B=22, \quad C=28-1-22 \Rightarrow C=5$
Now $B-2 A+C=22-2+5=25$
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