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If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{\left(x^2+1\right)}+\frac{B}{\left(x^2+1\right)^2}+$
$\frac{C}{\left(x^2+1\right)^3}$, then $A+C=$
Options:
$\frac{C}{\left(x^2+1\right)^3}$, then $A+C=$
Solution:
2741 Upvotes
Verified Answer
The correct answer is:
6
$$
\begin{aligned}
& \text { If } \frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2} \\
& +\frac{C}{\left(x^2+1\right)^3} \\
& \Rightarrow x^4+24 x^2+28=A\left(x^2+1\right)^2+B\left(x^2+1\right)+C \\
&
\end{aligned}
$$
On comparing the coefficient of different terms
$$
\begin{aligned}
& A=1 ; 2 A+B=24 \text { and } A+B+C=28 \\
& \Rightarrow \quad A=1, B=22 \text {, so } A+C=6 .
\end{aligned}
$$
\begin{aligned}
& \text { If } \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} \\
& \Rightarrow x^4+24 x^2+28=A\left(x^2+1\right)^2+B\left(x^2+1\right)+C \\
&
\end{aligned}
$$
On comparing the coefficient of different terms
$$
\begin{aligned}
& A=1 ; 2 A+B=24 \text { and } A+B+C=28 \\
& \Rightarrow \quad A=1, B=22 \text {, so } A+C=6 .
\end{aligned}
$$
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