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Integrate the function
$\int \frac{x^2+x+1}{(x+1)^2(x+2)} d x=I($ say $)$
$\int \frac{x^2+x+1}{(x+1)^2(x+2)} d x=I($ say $)$
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Now $\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+2}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}$
$\therefore \mathrm{x}^2+\mathrm{x}+1 \equiv \mathrm{A}(\mathrm{x}+1)^2+\mathrm{B}(\mathrm{x}+2)(\mathrm{x}+1)+\mathrm{C}(\mathrm{x}+2)$
Put $x=-2,-1, A=3, C=1$
Comparing the coefficient of $\mathrm{x}^2: \mathrm{B}=2$
$\begin{aligned}
&I=\int \frac{3}{x+2} d x-2 \int \frac{1}{x+1} d x+\int \frac{1}{(x+1)^2} d x \\
&=3 \log |x+2|-2 \log |x+1|-\frac{1}{x+1}+c
\end{aligned}$
$\therefore \mathrm{x}^2+\mathrm{x}+1 \equiv \mathrm{A}(\mathrm{x}+1)^2+\mathrm{B}(\mathrm{x}+2)(\mathrm{x}+1)+\mathrm{C}(\mathrm{x}+2)$
Put $x=-2,-1, A=3, C=1$
Comparing the coefficient of $\mathrm{x}^2: \mathrm{B}=2$
$\begin{aligned}
&I=\int \frac{3}{x+2} d x-2 \int \frac{1}{x+1} d x+\int \frac{1}{(x+1)^2} d x \\
&=3 \log |x+2|-2 \log |x+1|-\frac{1}{x+1}+c
\end{aligned}$
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