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

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Solution

$\int \frac{x^{2} + x + x + 1}{(x + 2) (x + 1)} \Rightarrow \frac{x^{2} + x + 1}{(x + 2) (x + 1)^{2}} = \frac{A}{(x + 2)} + \frac{B}{(x + 1)} + \frac{C}{(x + 1)} =$
$x^{2} + x + 1 = (x^{2} + 2 x + 1) A + B (x^{2} + 3 x + 12) + c (x + 2)$
$1 = A + B +$ $1 = 3 A + 4 B$
$1 = 2 A + 3 B + C) 2$ $(1 = A + B) 3$
$1 = A + 2 B + 2 C$ $3 = 3 A + 3 B$
$2 = 4 A + 2 B + 2 C$ __
_ $- 2 = B$
$1 = 3 A + 4 B$ $1 = 3 A - 8 \Rightarrow 3 A = 9$
$A = 3$

$- 3 + (- 4) + 2 C = 1$
$\int \frac{x^{2} + x + x + 1}{(x + 2) (x + 1)^{2}} = \int \frac{3}{(x + 2)} + \int \frac{- 2}{x + 1} + \int \frac{1}{(x + 1)^{2}}$
$= 3 l n (x + 2) - 2 ln (x + 1) - 9 x + 1) + C$

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