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

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

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

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

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