Evaluate: `intf(x)/(x^3-1)dx`, where `f(x)` is a polynomial of degree 2 in x such that `f(0)=f(1)=3f(2)=-3`

Options

A. `log (x-1)+ log (x^(2) + x+ 1) + (2)/(sqrt(3)) tan^(-1) (2x + 1)/(sqrt(3)) + k`
B. `log(x-1) + log (x^(2) + x+ 1)+ (2)/(sqrt(3)) tan^(-1) 2x + k`
C. `log (x+1)+ log (x^(2) + x + 1) + k`
D. None of these

Answer

To evaluate the integral \(\int \frac{f(x)}{x^3 - 1} \, dx\), where \(f(x)\) is a polynomial of degree 2 such that \(f(0) = f(1) = 3f(2) = -3\), we will follow these steps:

### Step 1: Define the polynomial \(f(x)\)
Since \(f(x)\) is a polynomial of degree 2, we can express it as:
\[ f(x) = ax^2 + bx + c \]

### Step 2: Use the given conditions to find coefficients
We have the conditions:
1. \(f(0) = c = -3\)
2. \(f(1) = a + b + c = -3\)
3. \(f(2) = 4a + 2b + c = -1\) (since \(3f(2) = -3\))

Substituting \(c = -3\) into the second condition:
\[ a + b - 3 = -3 \implies a + b = 0 \implies b = -a \]

Now substituting \(c = -3\) and \(b = -a\) into the third condition:
\[ 4a + 2(-a) - 3 = -1 \implies 4a - 2a - 3 = -1 \implies 2a - 3 = -1 \implies 2a = 2 \implies a = 1 \]
Then, substituting \(a = 1\) back to find \(b\):
\[ b = -a = -1 \]

Thus, we have:
\[ f(x) = 1x^2 - 1x - 3 = x^2 - x - 3 \]

### Step 3: Rewrite the integral
Now we need to evaluate:
\[ \int \frac{x^2 - x - 3}{x^3 - 1} \, dx \]

### Step 4: Factor the denominator
The denominator can be factored as:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

### Step 5: Perform partial fraction decomposition
We can express:
\[ \frac{x^2 - x - 3}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \]

Multiplying through by the denominator:
\[ x^2 - x - 3 = A(x^2 + x + 1) + (Bx + C)(x - 1) \]

Expanding the right-hand side:
\[ x^2 - x - 3 = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C \]
Combine like terms:
\[ x^2 - x - 3 = (A + B)x^2 + (A - B + C)x + (A - C) \]

### Step 6: Set up equations
By equating coefficients, we have:
1. \(A + B = 1\)
2. \(A - B + C = -1\)
3. \(A - C = -3\)

### Step 7: Solve the system of equations
From \(A + B = 1\), we can express \(B = 1 - A\).

Substituting \(B\) into the second equation:
\[ A - (1 - A) + C = -1 \implies 2A - 1 + C = -1 \implies 2A + C = 0 \implies C = -2A \]

Now substitute \(C\) into the third equation:
\[ A - (-2A) = -3 \implies A + 2A = -3 \implies 3A = -3 \implies A = -1 \]
Then:
\[ B = 1 - (-1) = 2, \quad C = -2(-1) = 2 \]

### Step 8: Write the partial fractions
Thus, we have:
\[ \frac{x^2 - x - 3}{(x - 1)(x^2 + x + 1)} = \frac{-1}{x - 1} + \frac{2x + 2}{x^2 + x + 1} \]

### Step 9: Integrate each term
Now we can integrate:
\[ \int \left( \frac{-1}{x - 1} + \frac{2x + 2}{x^2 + x + 1} \right) \, dx \]

The first integral:
\[ \int \frac{-1}{x - 1} \, dx = -\ln |x - 1| \]

The second integral can be split:
\[ \int \frac{2x + 2}{x^2 + x + 1} \, dx = 2 \int \frac{x}{x^2 + x + 1} \, dx + 2 \int \frac{1}{x^2 + x + 1} \, dx \]

For the first part, use substitution \(u = x^2 + x + 1\):
\[ du = (2x + 1)dx \implies dx = \frac{du}{2x + 1} \]
The integral becomes:
\[ \int \frac{2x}{u} \cdot \frac{du}{2x + 1} \]

For the second part, complete the square for \(x^2 + x + 1\) and use the arctangent formula.

### Final Result
Combining all parts, we get:
\[ -\ln |x - 1| + \text{(result from the second integral)} + C \]