Let y = frac{x^{2}}{(x+1)^{2}(x+2)}. Then fra | vedclass

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(A) Given $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$.Using partial fractions, we write:$\frac{x^{2}}{(x+1)^{2}(x+2)} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C

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$2\left[\frac{3}{(x+1)^{4}}-\frac{3}{(x+1)^{3}}+\frac{4}{(x+2)^{3}}\right]$

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(A) Given $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$.Using partial fractions, we write:$\frac{x^{2}}{(x+1)^{2}(x+2)} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C}{(x+1)^{2}}$.Solving for coefficients, we get $A=4, B=-3, C=1$.So, $y = 4(x+2)^{-1} - 3(x+1)^{-1} + (x+1)^{-2}$.Differentiating with respect to $x$ once:$y' = -4(x+2)^{-2} + 3(x+1)^{-2} - 2(x+1)^{-3}$.Differentiating again:$y'' = 8(x+2)^{-3} - 6(x+1)^{-3} + 6(x+1)^{-4}$.$y'' = 2\left[\frac{4}{(x+2)^{3}} - \frac{3}{(x+1)^{3}} + \frac{3}{(x+1)^{4}}\right]$.This matches option $A$.

Let $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$. Then $\frac{d^{2} y}{dx^{2}}$ is

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