Let's decompose the integrand into partial fractions:

$\frac{x^2 + x + 1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}$

Multiplying both sides by $(x+1)^2(x+2)$, we get:

$x^2 + x + 1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2$

Expanding the terms:

$x^2 + x + 1 = A(x^2 + 3x + 2) + B(x+2) + C(x^2 + 2x + 1)$

$x^2 + x + 1 = Ax^2 + 3Ax + 2A + Bx + 2B + Cx^2 + 2Cx + C$

Grouping like terms:

$x^2 + x + 1 = (A+C)x^2 + (3A+B+2C)x + (2A+2B+C)$

Comparing coefficients, we get the following system of equations:

1) $A + C = 1$

2) $3A + B + 2C = 1$

3) $2A + 2B + C = 1$

From equation (1), $C = 1 - A$. Substituting this into equations (2) and (3):

2) $3A + B + 2(1-A) = 1 \Rightarrow A + B = -1$

3) $2A + 2B + (1-A) = 1 \Rightarrow A + 2B = 0$

Solving the system of equations $A + B = -1$ and $A + 2B = 0$, we get:

Subtracting the first equation from the second, we get $B = 1$.

Substituting $B = 1$ into $A + B = -1$, we get $A = -2$.

Since $C = 1 - A$, we have $C = 1 - (-2) = 3$.

Thus, $A = -2$, $B = 1$, and $C = 3$.

So, $\frac{x^2 + x + 1}{(x+1)^2(x+2)} = \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2}$

Now, we integrate:

$\int \frac{x^2 + x + 1}{(x+1)^2(x+2)} dx = \int \left( \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} \right) dx$

$= -2 \int \frac{1}{x+1} dx + \int \frac{1}{(x+1)^2} dx + 3 \int \frac{1}{x+2} dx$

$= -2 \ln|x+1| - \frac{1}{x+1} + 3 \ln|x+2| + C$

Combining the logarithmic terms:

$= \ln\left| \frac{(x+2)^3}{(x+1)^2} \right| - \frac{1}{x+1} + C$