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

Answer

To evaluate the integral

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

we will use partial fraction decomposition and integration techniques.

### Step 1: Set up Partial Fraction Decomposition

We can express the integrand as:

\[ \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1} \]

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

### Step 2: Multiply through by the denominator

Multiplying both sides by \((x + 2)(x^2 + 1)\) gives:

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

### Step 3: Expand the right side

Expanding the right-hand side:

\[ A(x^2 + 1) + (Bx + C)(x + 2) = Ax^2 + A + Bx^2 + 2Bx + Cx + 2C. \]

Combining like terms, we have:

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

### Step 4: Set up equations for coefficients

Now, we equate coefficients from both sides:

1. For \(x^2\): \(A + B = 1\)
2. For \(x\): \(2B + C = 1\)
3. For the constant term: \(A + 2C = 1\)

### Step 5: Solve the system of equations

From the first equation \(A + B = 1\), we can express \(B\) in terms of \(A\):

\[ B = 1 - A. \]

Substituting \(B\) into the second equation:

\[ 2(1 - A) + C = 1 \implies 2 - 2A + C = 1 \implies C = 2A - 1. \]

Now substituting \(B\) and \(C\) into the third equation:

\[ A + 2(2A - 1) = 1 \implies A + 4A - 2 = 1 \implies 5A = 3 \implies A = \frac{3}{5}. \]

Using \(A\) to find \(B\) and \(C\):

\[ B = 1 - \frac{3}{5} = \frac{2}{5}, \]
\[ C = 2\left(\frac{3}{5}\right) - 1 = \frac{6}{5} - 1 = \frac{1}{5}. \]

### Step 6: Rewrite the integral

Now substituting back into the partial fractions:

\[ \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)} = \frac{3/5}{x + 2} + \frac{(2/5)x + (1/5)}{x^2 + 1}. \]

Thus, we can rewrite the integral as:

\[ I = \int \left( \frac{3/5}{x + 2} + \frac{(2/5)x + (1/5)}{x^2 + 1} \right) \, dx. \]

### Step 7: Integrate term by term

Now we can integrate each term separately:

1. For \(\frac{3/5}{x + 2}\):

\[ \int \frac{3/5}{x + 2} \, dx = \frac{3}{5} \ln |x + 2|. \]

2. For \(\frac{(2/5)x}{x^2 + 1}\):

Using the substitution \(u = x^2 + 1\), \(du = 2x \, dx\):

\[ \int \frac{(2/5)x}{x^2 + 1} \, dx = \frac{1}{5} \ln |x^2 + 1|. \]

3. For \(\frac{1/5}{x^2 + 1}\):

\[ \int \frac{1/5}{x^2 + 1} \, dx = \frac{1}{5} \tan^{-1}(x). \]

### Step 8: Combine results

Combining all parts, we have:

\[ I = \frac{3}{5} \ln |x + 2| + \frac{1}{5} \ln |x^2 + 1| + \frac{1}{5} \tan^{-1}(x) + C, \]

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

### Final Answer

Thus, the evaluated integral is:

\[ I = \frac{3}{5} \ln |x + 2| + \frac{1}{5} \ln |x^2 + 1| + \frac{1}{5} \tan^{-1}(x) + C. \]