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If $\frac{-x^2+6 x+1}{(x-1)^2\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x-3}{x^2+2}$, then $\mathrm{A}+\mathrm{B}+\mathrm{C}=$
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Verified Answer
The correct answer is:
$2$
Given,
$\begin{aligned}
& \frac{-x^2+6 x+1}{(x-1)^2\left(x^2+2\right)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C x-3}{x^2+2} \\
& \Rightarrow A(x-1)\left(x^2+2\right)+B\left(x^2+2\right)+(C x-3) \times \\
& (x-1)^2=-x^2+6 x+1 \\
& \text { If } x=1 \Rightarrow 3 B=6 \Rightarrow B=2 \\
& \text { and } A+C=0 \\
& \text { Now, } A+B+C=2
\end{aligned}$
$\begin{aligned}
& \frac{-x^2+6 x+1}{(x-1)^2\left(x^2+2\right)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C x-3}{x^2+2} \\
& \Rightarrow A(x-1)\left(x^2+2\right)+B\left(x^2+2\right)+(C x-3) \times \\
& (x-1)^2=-x^2+6 x+1 \\
& \text { If } x=1 \Rightarrow 3 B=6 \Rightarrow B=2 \\
& \text { and } A+C=0 \\
& \text { Now, } A+B+C=2
\end{aligned}$
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