If frac{x+3}{(x+1)(x^2+2)} = frac{a}{x+1} + f | vedclass

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(C) Given the partial fraction decomposition: $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$.Multiplying both sides by $(x+1)(x^2+2)$

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$3$

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(C) Given the partial fraction decomposition: $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$.Multiplying both sides by $(x+1)(x^2+2)$, we get: $x+3 = a(x^2+2) + (bx+c)(x+1)$.Expanding the right side: $x+3 = ax^2 + 2a + bx^2 + bx + cx + c$.Grouping terms by powers of $x$: $x+3 = (a+b)x^2 + (b+c)x + (2a+c)$.Comparing coefficients on both sides:$1$) $a+b = 0 \implies b = -a$$2$) $b+c = 1$$3$) $2a+c = 3$Substitute $b = -a$ into $(2)$: $-a+c = 1 \implies c = a+1$.Substitute $c = a+1$ into $(3)$: $2a + (a+1) = 3 \implies 3a = 2 \implies a = \frac{2}{3}$.Then $b = -\frac{2}{3}$ and $c = \frac{2}{3} + 1 = \frac{5}{3}$.Finally, calculate $a-b+c = \frac{2}{3} - (-\frac{2}{3}) + \frac{5}{3} = \frac{2}{3} + \frac{2}{3} + \frac{5}{3} = \frac{9}{3} = 3$.

If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$, then $a-b+c=$

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