If frac{x^3+x^2+1}{(x^2+2)(x^2+3)}=frac{Ax+B} | vedclass

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(D) Given the partial fraction decomposition: $\frac{x^3+x^2+1}{(x^2+2)(x^2+3)}=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+3}$.Multiplying both sides by $(x^2

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(D) Given the partial fraction decomposition: $\frac{x^3+x^2+1}{(x^2+2)(x^2+3)}=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+3}$.Multiplying both sides by $(x^2+2)(x^2+3)$,we get: $x^3+x^2+1 = (Ax+B)(x^2+3) + (Cx+D)(x^2+2)$.Expanding the right side: $x^3+x^2+1 = Ax^3 + 3Ax + Bx^2 + 3B + Cx^3 + 2Cx + Dx^2 + 2D$.Grouping the terms by powers of $x$: $x^3+x^2+1 = (A+C)x^3 + (B+D)x^2 + (3A+2C)x + (3B+2D)$.Comparing coefficients on both sides:$1$) $A+C = 1$$2$) $B+D = 1$$3$) $3A+2C = 0$$4$) $3B+2D = 1$From $(1)$,$C = 1-A$. Substituting into $(3)$: $3A + 2(1-A) = 0 \implies A+2=0 \implies A=-2$. Then $C = 1-(-2) = 3$.From $(2)$,$D = 1-B$. Substituting into $(4)$: $3B + 2(1-B) = 1 \implies B+2=1 \implies B=-1$. Then $D = 1-(-1) = 2$.Thus,$A+B+C+D = -2 - 1 + 3 + 2 = 2$.

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

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