If the integral `int(x^(4)+x^(2)+1)/(x^(2)x-x+1)dx=f(x)+C,` (where C is the constant of integration and `x in R`), then the minimum value of `f'(x)` is

Options

A. 1
B. `(1)/(4)`
C. `(3)/(4)`
D. 2

Answer

To solve the problem, we need to find the minimum value of \( f'(x) \) given that

\[ \int \frac{x^4 + x^2 + 1}{x^2 x - x + 1} \, dx = f(x) + C \]

### Step 1: Simplify the integrand

First, we simplify the integrand:

\[ \frac{x^4 + x^2 + 1}{x^3 - x + 1} \]

We can analyze the numerator and denominator to see if we can simplify or factor them.

### Step 2: Perform polynomial long division (if necessary)

If the degree of the numerator is greater than or equal to the degree of the denominator, we perform polynomial long division. Here, the degree of the numerator (4) is greater than the degree of the denominator (3).

1. Divide \( x^4 \) by \( x^3 \) to get \( x \).
2. Multiply \( x \) by the entire denominator \( (x^3 - x + 1) \) to get \( x^4 - x^2 + x \).
3. Subtract this from the numerator:

\[ (x^4 + x^2 + 1) - (x^4 - x^2 + x) = 2x^2 - x + 1 \]

So we can rewrite the integrand as:

\[ x + \frac{2x^2 - x + 1}{x^3 - x + 1} \]

### Step 3: Integrate the simplified expression

Now we can integrate:

\[ \int \left( x + \frac{2x^2 - x + 1}{x^3 - x + 1} \right) \, dx \]

This gives us:

\[ \int x \, dx + \int \frac{2x^2 - x + 1}{x^3 - x + 1} \, dx \]

The first integral is straightforward:

\[ \frac{x^2}{2} + C_1 \]

### Step 4: Find \( f'(x) \)

To find \( f'(x) \), we differentiate \( f(x) \):

\[ f'(x) = x + \frac{d}{dx} \left( \int \frac{2x^2 - x + 1}{x^3 - x + 1} \, dx \right) \]

Using the Fundamental Theorem of Calculus, we have:

\[ f'(x) = x + \frac{2x^2 - x + 1}{x^3 - x + 1} \]

### Step 5: Analyze \( f'(x) \)

Now we need to find the minimum value of:

\[ f'(x) = x + \frac{2x^2 - x + 1}{x^3 - x + 1} \]

### Step 6: Find critical points

To find the minimum value, we can set the derivative \( f''(x) \) to zero and solve for \( x \):

1. Differentiate \( f'(x) \) to find \( f''(x) \).
2. Set \( f''(x) = 0 \) and solve for \( x \).

### Step 7: Completing the square

To find the minimum value of the quadratic part \( 2x^2 - x + 1 \):

1. Rewrite it in the form \( 2\left(x - \frac{1}{4}\right)^2 + \frac{3}{4} \).
2. The minimum value occurs at \( x = \frac{1}{4} \) and is \( \frac{3}{4} \).

### Conclusion

Thus, the minimum value of \( f'(x) \) is:

\[ \frac{3}{4} \]