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

\(
I=\int \frac{x^{2}}{(x+1)(x+2)^{2}} d x
\)
\(
\begin{array}{l}
\text {Let } \frac{\mathrm{x}^{2}}{(\mathrm{x}+1)(\mathrm{x}+2)^{2}}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}+\frac{\mathrm{C}}{(\mathrm{x}+2)^{2}} \\
\therefore \mathrm{x}^{2}=\mathrm{A}(\mathrm{x}+2)^{2}+\mathrm{B}(\mathrm{x}+1)(\mathrm{x}+2)+\mathrm{C}(\mathrm{x}+1) \\
\text {When } \mathrm{x}=-2 \text {, we get } 4=-\mathrm{C} \Rightarrow \mathrm{C}=-4 \\
\text {When } \mathrm{x}=-1 \text {,we get } 1=\mathrm{A} \Rightarrow \mathrm{A}=1
\end{array}
\)
When \(x=0, A=1, C=-4\), we get \(0=4+2 B-4 \Rightarrow B=0\)
\(\therefore I =\int\left[\frac{1}{x+1}-\frac{4}{(x+2)^{2}}\right] d x \)
\( =\int \frac{d x}{x+1}-4 \int(x+2)^{-2} d x \)
\( =\log |x+1|-4 \frac{(x+2)^{-1}}{(-1)}+c \)
\( =\log |x+1|+\frac{4}{x+2}+c\)