If the integral `int(x^(4)+x^(2)+1)/(x^(2)-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 integral

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

we will first simplify the integrand.

### Step 1: Simplify the integrand

We can express the numerator \(x^4 + x^2 + 1\) in terms of the denominator \(x^2 - x + 1\). We can try polynomial long division or factorization.

Notice that:

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

Thus, we can rewrite the integral as:

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

This simplifies to:

\[ \int (x^2 + x + 1) \, dx \]

### Step 2: Integrate the simplified expression

Now we can integrate:

\[ \int (x^2 + x + 1) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C \]

So we have:

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

### Step 3: Differentiate \(f(x)\)

Next, we differentiate \(f(x)\):

\[ f'(x) = x^2 + x + 1 \]

### Step 4: Find the minimum value of \(f'(x)\)

To find the minimum value of \(f'(x)\), we can analyze the quadratic function \(f'(x) = x^2 + x + 1\).

The vertex of a quadratic function \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = 1\):

\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \]

### Step 5: Evaluate \(f'(-\frac{1}{2})\)

Now we substitute \(x = -\frac{1}{2}\) back into \(f'(x)\):

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

Calculating this:

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

### Conclusion

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

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