If frac{A}{x-a}+frac{B x+C}{x^2+b^2}=frac{1}{ | vedclass

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(C) Given the equation: $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)(x^2+b^2)}$Multiplying both sides by $(x-a)(x^2+b^2)$,we get:$A(x^2+b^2)+(B

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$\frac{-a}{a^2+b^2}$

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(C) Given the equation: $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)(x^2+b^2)}$Multiplying both sides by $(x-a)(x^2+b^2)$,we get:$A(x^2+b^2)+(B x+C)(x-a)=1$Expanding the terms:$(A+B)x^2+(C-a B)x+(A b^2-a C)=1$Comparing the coefficients of $x^2$,$x$,and the constant term on both sides:$A+B=0$  ....$(i)$$C-a B=0$  ....$(ii)$$A b^2-a C=1$  ....$(iii)$From $(i)$,$B=-A$. Substituting this into $(ii)$:$C-a(-A)=0 \Rightarrow C+a A=0 \Rightarrow A=\frac{-C}{a}$Substituting $A=\frac{-C}{a}$ into $(iii)$:$(\frac{-C}{a})b^2-a C=1$$-C(\frac{b^2+a^2}{a})=1$$C=\frac{-a}{a^2+b^2}$

If $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)(x^2+b^2)}$ then $C=$

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