If frac{d}{d x}left(frac{x^2}{(x+2)(2 x+3)}ri | vedclass

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(B) Let $f(x) = \frac{x^2}{(x+2)(2x+3)} = \frac{x^2}{2x^2+7x+6}$.We perform partial fraction decomposition on $f(x)$ first:$\frac{x^2}{(x+2)(2x+3)} =

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$-5$

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(B) Let $f(x) = \frac{x^2}{(x+2)(2x+3)} = \frac{x^2}{2x^2+7x+6}$.We perform partial fraction decomposition on $f(x)$ first:$\frac{x^2}{(x+2)(2x+3)} = \frac{1}{2} + \frac{-\frac{7}{2}x - 3}{(x+2)(2x+3)} = \frac{1}{2} + \frac{A'}{x+2} + \frac{B'}{2x+3}$.Using partial fractions: $\frac{-\frac{7}{2}x - 3}{(x+2)(2x+3)} = \frac{P}{x+2} + \frac{Q}{2x+3}$.Setting $x = -2$: $P = \frac{-\frac{7}{2}(-2) - 3}{2(-2)+3} = \frac{4}{-1} = -4$.Setting $x = -3/2$: $Q = \frac{-\frac{7}{2}(-3/2) - 3}{(-3/2)+2} = \frac{21/4 - 12/4}{1/2} = \frac{9/4}{1/2} = 9/2$.So,$f(x) = \frac{1}{2} - \frac{4}{x+2} + \frac{9/2}{2x+3}$.Now,differentiate with respect to $x$:$\frac{d}{dx} f(x) = 0 - 4(-1)(x+2)^{-2} + \frac{9}{2}(-1)(2x+3)^{-2} \cdot 2$$\frac{d}{dx} f(x) = \frac{4}{(x+2)^2} - \frac{9}{(2x+3)^2}$.Comparing this with $\frac{A}{(x+2)^2} + \frac{B}{(2x+3)^2}$,we get $A = 4$ and $B = -9$.Therefore,$A+B = 4 + (-9) = -5$.

If $\frac{d}{d x}\left(\frac{x^2}{(x+2)(2 x+3)}\right)=\frac{A}{(x+2)^2}+\frac{B}{(2 x+3)^2}$ then $A+B=$

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