If frac{d}{{dx}}left[ {frac{{2{x^3} + 3{x^2} | vedclass

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(C) First,simplify the expression inside the derivative using polynomial division and partial fractions.Let $f(x) = \frac{2x^3 + 3x^2 + x - 3}{x^2 + x

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$-2$

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(C) First,simplify the expression inside the derivative using polynomial division and partial fractions.Let $f(x) = \frac{2x^3 + 3x^2 + x - 3}{x^2 + x - 2}$.Since $x^2 + x - 2 = (x - 1)(x + 2)$,we perform division:$2x^3 + 3x^2 + x - 3 = (2x + 1)(x^2 + x - 2) + 0x - 1$,which is not quite right. Let's re-evaluate:$\frac{2x^3 + 3x^2 + x - 3}{x^2 + x - 2} = 2x + 1 + \frac{-2x - 1}{x^2 + x - 2}$.Using partial fractions for $\frac{-2x - 1}{(x - 1)(x + 2)} = \frac{P}{x - 1} + \frac{Q}{x + 2}$.$-2x - 1 = P(x + 2) + Q(x - 1)$.For $x = 1$: $-3 = 3P \implies P = -1$.For $x = -2$: $3 = -3Q \implies Q = -1$.So,$f(x) = 2x + 1 - \frac{1}{x - 1} - \frac{1}{x + 2}$.Now,differentiate with respect to $x$:$\frac{d}{dx}f(x) = 2 + \frac{1}{(x - 1)^2} + \frac{1}{(x + 2)^2}$.Comparing this with $A + \frac{B}{(x - 1)^2} + \frac{C}{(x + 2)^2}$,we get $A = 2$,$B = 1$,$C = 1$.Therefore,$A - B + C = 2 - 1 + 1 = 2$.

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

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