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

Answer

To evaluate the integral

\[ I = \int \frac{x^2 + x + 1}{x^2 (x + 2)} \, dx, \]

we will use the method of partial fractions.

### Step 1: Set up the partial fraction decomposition

We can express the integrand as:

\[ \frac{x^2 + x + 1}{x^2 (x + 2)} = \frac{Ax + B}{x^2} + \frac{C}{x + 2}, \]

where \(A\), \(B\), and \(C\) are constants to be determined.

### Step 2: Combine the right-hand side

Multiplying both sides by the denominator \(x^2 (x + 2)\) gives:

\[ x^2 + x + 1 = (Ax + B)(x + 2) + Cx^2. \]

Expanding the right-hand side:

\[ (Ax + B)(x + 2) = Ax^2 + 2Ax + Bx + 2B = Ax^2 + (2A + B)x + 2B. \]

Thus, we have:

\[ x^2 + x + 1 = (A + C)x^2 + (2A + B)x + 2B. \]

### Step 3: Compare coefficients

Now, we can compare coefficients from both sides:

1. Coefficient of \(x^2\): \(A + C = 1\)
2. Coefficient of \(x\): \(2A + B = 1\)
3. Constant term: \(2B = 1\)

### Step 4: Solve the system of equations

From the third equation \(2B = 1\), we find:

\[ B = \frac{1}{2}. \]

Substituting \(B\) into the second equation:

\[ 2A + \frac{1}{2} = 1 \implies 2A = \frac{1}{2} \implies A = \frac{1}{4}. \]

Now substituting \(A\) into the first equation:

\[ \frac{1}{4} + C = 1 \implies C = 1 - \frac{1}{4} = \frac{3}{4}. \]

### Step 5: Substitute back into the integral

Now we have:

\[ A = \frac{1}{4}, \quad B = \frac{1}{2}, \quad C = \frac{3}{4}. \]

Thus, we can rewrite the integral as:

\[ I = \int \left( \frac{1/4 \cdot x + 1/2}{x^2} + \frac{3/4}{x + 2} \right) \, dx. \]

### Step 6: Separate the integral

This can be separated into three integrals:

\[ I = \frac{1}{4} \int \frac{x}{x^2} \, dx + \frac{1}{2} \int \frac{1}{x^2} \, dx + \frac{3}{4} \int \frac{1}{x + 2} \, dx. \]

### Step 7: Evaluate each integral

1. \(\int \frac{x}{x^2} \, dx = \int \frac{1}{x} \, dx = \log |x|\)
2. \(\int \frac{1}{x^2} \, dx = -\frac{1}{x}\)
3. \(\int \frac{1}{x + 2} \, dx = \log |x + 2|\)

### Step 8: Combine the results

Now substituting back, we get:

\[ I = \frac{1}{4} \log |x| - \frac{1}{2} \cdot \frac{1}{x} + \frac{3}{4} \log |x + 2| + C, \]

where \(C\) is the constant of integration.

### Final Result

Thus, the final result of the integral is:

\[ I = \frac{1}{4} \log |x| - \frac{1}{2x} + \frac{3}{4} \log |x + 2| + C. \]

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