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

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(B) Let $y = x^2$. The expression becomes $\frac{y+1}{(y+2)(y+3)} = \frac{Ay+B}{y+2} + \frac{Cy+D}{y+3}$.Using partial fractions for $\frac{y+1}{(y+2)

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(B) Let $y = x^2$. The expression becomes $\frac{y+1}{(y+2)(y+3)} = \frac{Ay+B}{y+2} + \frac{Cy+D}{y+3}$.Using partial fractions for $\frac{y+1}{(y+2)(y+3)}$, we have $\frac{y+1}{(y+2)(y+3)} = \frac{P}{y+2} + \frac{Q}{y+3}$.$y+1 = P(y+3) + Q(y+2)$.For $y = -2$, $-2+1 = P(-2+3) \implies P = -1$.For $y = -3$, $-3+1 = Q(-3+2) \implies -2 = -Q \implies Q = 2$.So, $\frac{y+1}{(y+2)(y+3)} = \frac{-1}{y+2} + \frac{2}{y+3}$.Substituting back $y = x^2$, we get $\frac{-1}{x^2+2} + \frac{2}{x^2+3}$.Comparing this with $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$, we get $A=0, B=-1, C=0, D=2$.Therefore, $A+B+C+D = 0 + (-1) + 0 + 2 = 1$.

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

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