int frac{x^{2}}{(x+1)(x+2)^{2}} d x= | vedclass

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(A) Let $I = \int \frac{x^{2}}{(x+1)(x+2)^{2}} dx$. Using partial fractions,we write: $\frac{x^{2}}{(x+1)(x+2)^{2}} = \frac{A}{x+1} + \frac{B}{x+2} +

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$\log |x+1|+\frac{4}{x+2}+c$

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(A) Let $I = \int \frac{x^{2}}{(x+1)(x+2)^{2}} dx$. Using partial fractions,we write: $\frac{x^{2}}{(x+1)(x+2)^{2}} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^{2}}$. Multiplying both sides by $(x+1)(x+2)^{2}$,we get: $x^{2} = A(x+2)^{2} + B(x+1)(x+2) + C(x+1)$. Setting $x = -2$,we get $(-2)^{2} = C(-2+1) \Rightarrow 4 = -C \Rightarrow C = -4$. Setting $x = -1$,we get $(-1)^{2} = A(-1+2)^{2} \Rightarrow 1 = A(1)^{2} \Rightarrow A = 1$. Comparing the coefficients of $x^{2}$ on both sides,we get $1 = A + B$. Since $A = 1$,we have $1 = 1 + B \Rightarrow B = 0$. Thus,the integral becomes: $I = \int \left( \frac{1}{x+1} - \frac{4}{(x+2)^{2}} \right) dx$. $I = \int \frac{1}{x+1} dx - 4 \int (x+2)^{-2} dx$. $I = \log |x+1| - 4 \left( \frac{(x+2)^{-1}}{-1} \right) + c$. $I = \log |x+1| + \frac{4}{x+2} + c$.

$\int \frac{x^{2}}{(x+1)(x+2)^{2}} d x=$

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