Let f(x) be a polynomial of degree three `f(0) = -1 and f(1) = 0.` Also, 0 is a stationary point of `f(x).` If f(x) does not have an extremum at `x=0,` then the value of integral `int(f(x))/(x^3-1)dx,` is

Options

A. `(x^(2))/(2)+C`
B. x+C
C. `(x^(3))/(6)+C`
D. nome of these

Explanation

To solve the problem, we need to find the value of the integral

\[ \int \frac{f(x)}{x^3 - 1} \, dx \]

given the conditions on the polynomial \( f(x) \).

### Step 1: Define the polynomial \( f(x) \)

Since \( f(x) \) is a polynomial of degree 3, we can express it in the general form:

\[ f(x) = ax^3 + bx^2 + cx + d \]

### Step 2: Use the given conditions

We have the following conditions:
1. \( f(0) = -1 \)
2. \( f(1) = 0 \)
3. \( 0 \) is a stationary point, meaning \( f'(0) = 0 \).

From \( f(0) = -1 \):

\[ f(0) = d = -1 \]

So, we can rewrite \( f(x) \) as:

\[ f(x) = ax^3 + bx^2 + cx - 1 \]

### Step 3: Apply the second condition \( f(1) = 0 \)

Now, substituting \( x = 1 \):

\[ f(1) = a(1)^3 + b(1)^2 + c(1) - 1 = 0 \]

This simplifies to:

\[ a + b + c - 1 = 0 \quad \Rightarrow \quad a + b + c = 1 \quad \text{(Equation 1)} \]

### Step 4: Use the stationary point condition

The derivative of \( f(x) \) is:

\[ f'(x) = 3ax^2 + 2bx + c \]

Setting \( f'(0) = 0 \):

\[ f'(0) = c = 0 \]

Now substituting \( c = 0 \) into Equation 1:

\[ a + b + 0 = 1 \quad \Rightarrow \quad a + b = 1 \quad \text{(Equation 2)} \]

### Step 5: Determine the coefficients

Now we have two equations:
1. \( c = 0 \)
2. \( a + b = 1 \)

We can express \( b \) in terms of \( a \):

\[ b = 1 - a \]

Thus, we can express \( f(x) \) as:

\[ f(x) = ax^3 + (1 - a)x^2 - 1 \]

### Step 6: Choose a specific value for \( a \)

Since we need \( f(x) \) to not have an extremum at \( x = 0 \), we can choose \( a = 1 \) (as an example). Then:

\[ b = 1 - 1 = 0 \]

Thus, we have:

\[ f(x) = x^3 - 1 \]

### Step 7: Evaluate the integral

Now we need to evaluate:

\[ \int \frac{f(x)}{x^3 - 1} \, dx = \int \frac{x^3 - 1}{x^3 - 1} \, dx \]

This simplifies to:

\[ \int 1 \, dx = x + C \]

### Final Answer

The value of the integral is:

\[ \int \frac{f(x)}{x^3 - 1} \, dx = x + C \]