`int((x^(2)+x+1))/((x+2)(x+1)^(2))dx`

Answer

To solve the integral \( \int \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} \, dx \), we will use the method of partial fractions. Let's go through the steps systematically.

### Step 1: Set up the partial fraction decomposition
We can express the integrand as:
\[ \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} = \frac{A}{x + 2} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2} \]
where \( A \), \( B \), and \( C \) are constants we need to determine.

### Step 2: Clear the denominators
Multiply both sides by the denominator \((x + 2)(x + 1)^2\):
\[ x^2 + x + 1 = A(x + 1)^2 + B(x + 2)(x + 1) + C(x + 2) \]

### Step 3: Expand the right-hand side
Expanding the right-hand side:
\[ A(x^2 + 2x + 1) + B(x^2 + 3x + 2) + C(x + 2) \]
This simplifies to:
\[ (A + B)x^2 + (2A + 3B + C)x + (A + 2B + 2C) \]

### Step 4: Set up the system of equations
Now, we can equate coefficients from both sides:
1. For \( x^2 \): \( A + B = 1 \)
2. For \( x \): \( 2A + 3B + C = 1 \)
3. For the constant term: \( A + 2B + 2C = 1 \)

### Step 5: Solve the system of equations
From equation (1): \( A + B = 1 \) implies \( B = 1 - A \).

Substituting \( B \) into equations (2) and (3):
- From (2):
\[ 2A + 3(1 - A) + C = 1 \implies 2A + 3 - 3A + C = 1 \implies -A + C = -2 \implies C = A - 2 \]

- From (3):
\[ A + 2(1 - A) + 2(A - 2) = 1 \implies A + 2 - 2A + 2A - 4 = 1 \implies A - 2 = 1 \implies A = 3 \]
Substituting \( A = 3 \) back to find \( B \) and \( C \):
\[ B = 1 - 3 = -2 \]
\[ C = 3 - 2 = 1 \]

### Step 6: Write the partial fractions
Now we have:
\[ \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} = \frac{3}{x + 2} - \frac{2}{x + 1} + \frac{1}{(x + 1)^2} \]

### Step 7: Integrate each term
Now we can integrate each term separately:
\[ \int \left( \frac{3}{x + 2} - \frac{2}{x + 1} + \frac{1}{(x + 1)^2} \right) \, dx \]

1. \( \int \frac{3}{x + 2} \, dx = 3 \ln |x + 2| + C_1 \)
2. \( \int -\frac{2}{x + 1} \, dx = -2 \ln |x + 1| + C_2 \)
3. \( \int \frac{1}{(x + 1)^2} \, dx = -\frac{1}{x + 1} + C_3 \)

### Step 8: Combine the results
Combining all the integrals, we get:
\[ \int \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} \, dx = 3 \ln |x + 2| - 2 \ln |x + 1| - \frac{1}{x + 1} + C \]

### Final Answer
\[ \int \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} \, dx = 3 \ln |x + 2| - 2 \ln |x + 1| - \frac{1}{x + 1} + C \]