If the partial fraction decomposition of frac | vedclass

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(B) Given the expression $\frac{x^2+1}{x^3+3x^2+3x+2} = \frac{x^2+1}{(x+2)(x^2+x+1)}$.We write the partial fraction decomposition as $\frac{x^2+1}{(x+

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$2$

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(B) Given the expression $\frac{x^2+1}{x^3+3x^2+3x+2} = \frac{x^2+1}{(x+2)(x^2+x+1)}$.We write the partial fraction decomposition as $\frac{x^2+1}{(x+2)(x^2+x+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$.Multiplying both sides by $(x+2)(x^2+x+1)$,we get $x^2+1 = A(x^2+x+1) + (Bx+C)(x+2)$.Setting $x = -2$: $(-2)^2 + 1 = A(4-2+1)$ $\Rightarrow 5 = 3A$ $\Rightarrow A = \frac{5}{3}$.Comparing the coefficients of $x^2$: $1 = A + B \Rightarrow B = 1 - \frac{5}{3} = -\frac{2}{3}$.Comparing the constant terms: $1 = A + 2C$ $\Rightarrow 1 = \frac{5}{3} + 2C$ $\Rightarrow 2C = 1 - \frac{5}{3} = -\frac{2}{3}$ $\Rightarrow C = -\frac{1}{3}$.Therefore,$A - B + C = \frac{5}{3} - (-\frac{2}{3}) + (-\frac{1}{3}) = \frac{5+2-1}{3} = \frac{6}{3} = 2$.

If the partial fraction decomposition of $\frac{x^2+1}{x^3+3x^2+3x+2}$ is $\frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$,then find the value of $A-B+C$. Note: The original expression provided in the prompt was corrected to the standard form $\frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$.

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