`int (4x^(2) +x+1)/(x^(3) -1) dx=?`

Answer

To solve the integral \( \int \frac{4x^2 + x + 1}{x^3 - 1} \, dx \), we can follow these steps:

### Step 1: Factor the Denominator
The denominator \( x^3 - 1 \) can be factored using the formula for the difference of cubes:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

### Step 2: Rewrite the Integral
Now, we can rewrite the integral as:
\[ \int \frac{4x^2 + x + 1}{(x - 1)(x^2 + x + 1)} \, dx \]

### Step 3: Partial Fraction Decomposition
We can express the integrand using partial fractions:
\[ \frac{4x^2 + x + 1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \]
where \( A \), \( B \), and \( C \) are constants to be determined.

### Step 4: Clear the Denominator
Multiply through by the denominator \( (x - 1)(x^2 + x + 1) \):
\[ 4x^2 + x + 1 = A(x^2 + x + 1) + (Bx + C)(x - 1) \]

### Step 5: Expand and Combine Like Terms
Expanding the right-hand side:
\[ 4x^2 + x + 1 = A(x^2 + x + 1) + Bx^2 - Bx + Cx - C \]
Combine like terms:
\[ 4x^2 + x + 1 = (A + B)x^2 + (A - B + C)x + (A - C) \]

### Step 6: Set Up the System of Equations
Now, we can set up a system of equations by comparing coefficients:
1. \( A + B = 4 \) (coefficient of \( x^2 \))
2. \( A - B + C = 1 \) (coefficient of \( x \))
3. \( A - C = 1 \) (constant term)

### Step 7: Solve the System of Equations
From equation 3, we can express \( C \) in terms of \( A \):
\[ C = A - 1 \]
Substituting \( C \) into equation 2:
\[ A - B + (A - 1) = 1 \implies 2A - B - 1 = 1 \implies 2A - B = 2 \implies B = 2A - 2 \]
Now substituting \( B \) into equation 1:
\[ A + (2A - 2) = 4 \implies 3A - 2 = 4 \implies 3A = 6 \implies A = 2 \]
Now substituting \( A \) back to find \( B \) and \( C \):
\[ B = 2(2) - 2 = 2 \]
\[ C = 2 - 1 = 1 \]

### Step 8: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \left( \frac{2}{x - 1} + \frac{2x + 1}{x^2 + x + 1} \right) \, dx \]

### Step 9: Integrate Each Term
Now we can integrate each term separately:
1. \( \int \frac{2}{x - 1} \, dx = 2 \ln |x - 1| \)
2. For \( \int \frac{2x + 1}{x^2 + x + 1} \, dx \), we can use substitution:
Let \( t = x^2 + x + 1 \), then \( dt = (2x + 1) \, dx \). Thus, \( \int \frac{2x + 1}{x^2 + x + 1} \, dx = \ln |t| + C = \ln |x^2 + x + 1| + C \).

### Step 10: Combine the Results
Combining the results, we have:
\[ \int \frac{4x^2 + x + 1}{x^3 - 1} \, dx = 2 \ln |x - 1| + \ln |x^2 + x + 1| + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{4x^2 + x + 1}{x^3 - 1} \, dx = 2 \ln |x - 1| + \ln |x^2 + x + 1| + C \]