text{If } frac{3x^2+1}{(x^2+1)(x^2+2)^2} = fr | vedclass

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(A) Given the partial fraction decomposition: $\frac{3x^2+1}{(x^2+1)(x^2+2)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2} + \frac{Ex+F}{(x^2+2)^2}$ Mul

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(A) Given the partial fraction decomposition: $\frac{3x^2+1}{(x^2+1)(x^2+2)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2} + \frac{Ex+F}{(x^2+2)^2}$ Multiplying both sides by the denominator $(x^2+1)(x^2+2)^2$,we get: $3x^2+1 = (Ax+B)(x^2+2)^2 + (Cx+D)(x^2+1)(x^2+2) + (Ex+F)(x^2+1)$ Since the expression $3x^2+1$ contains only even powers of $x$,the coefficients of all odd powers of $x$ (i.e.,$x^5, x^3, x^1$) must be zero. Expanding the terms: $(Ax+B)(x^4+4x^2+4) = Ax^5 + 4Ax^3 + 4Ax + Bx^4 + 4Bx^2 + 4B$ $(Cx+D)(x^4+3x^2+2) = Cx^5 + 3Cx^3 + 2Cx + Dx^4 + 3Dx^2 + 2D$ $(Ex+F)(x^2+1) = Ex^3 + Ex + Fx^2 + F$ Equating the coefficients of $x^5$: $A + C = 0$ Equating the coefficients of $x^3$: $4A + 3C + E = 0$ Since $A+C=0$,we have $C = -A$. Substituting this into the second equation: $4A + 3(-A) + E = 0 \Rightarrow A + E = 0$ Thus,$A=0, C=0, E=0$. Therefore,$A+C+E = 0+0+0 = 0$.

$\text{If } \frac{3x^2+1}{(x^2+1)(x^2+2)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2} + \frac{Ex+F}{(x^2+2)^2}, \text{ then } A+C+E = $

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