If frac{d}{d x}left(frac{2 x+1}{(x+1)^2(x-2)} | vedclass

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(C) Let $y = \frac{2x+1}{(x+1)^2(x-2)}$. Using partial fractions, we write: $\frac{2x+1}{(x+1)^2(x-2)} = \frac{a}{x-2} + \frac{b}{x+1} + \frac{c}{(x+1

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$-2/3$

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(C) Let $y = \frac{2x+1}{(x+1)^2(x-2)}$. Using partial fractions, we write: $\frac{2x+1}{(x+1)^2(x-2)} = \frac{a}{x-2} + \frac{b}{x+1} + \frac{c}{(x+1)^2}$.
Solving for constants: $2x+1 = a(x+1)^2 + b(x+1)(x-2) + c(x-2)$.
For $x=2$: $5 = a(3)^2 \Rightarrow a = \frac{5}{9}$.
For $x=-1$: $-1 = c(-3) \Rightarrow c = \frac{1}{3}$.
Comparing coefficients of $x^2$: $0 = a + b \Rightarrow b = -a = -\frac{5}{9}$.
So, $y = \frac{5/9}{x-2} - \frac{5/9}{x+1} + \frac{1/3}{(x+1)^2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -\frac{5/9}{(x-2)^2} + \frac{5/9}{(x+1)^2} - \frac{2/3}{(x+1)^3}$.
Comparing this with $\frac{A}{(x-2)^2} + \frac{B}{(x+1)^3} + \frac{C}{(x+1)^2}$, we find $A = -\frac{5}{9}$, $B = -\frac{2}{3}$, and $C = \frac{5}{9}$.
Therefore, $A+B+C = -\frac{5}{9} - \frac{2}{3} + \frac{5}{9} = -\frac{2}{3}$.

If $\frac{d}{d x}\left(\frac{2 x+1}{(x+1)^2(x-2)}\right)=\frac{A}{(x-2)^2}+\frac{B}{(x+1)^3}+\frac{C}{(x+1)^2}$, then $A+B+C=$

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