`int_(-1)^(1) (sqrt(1+x+x^(2))-sqrt(1-x+x^(2))) dx =`

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the definite integral

\[ I = \int_{-1}^{1} \left( \sqrt{1+x+x^2} - \sqrt{1-x+x^2} \right) \, dx, \]

we will analyze the integrand to determine if it is an odd function.

### Step 1: Define the function
Let

\[ f(x) = \sqrt{1+x+x^2} - \sqrt{1-x+x^2}. \]

### Step 2: Check for oddness
To check if \( f(x) \) is an odd function, we need to compute \( f(-x) \):

\[ f(-x) = \sqrt{1 + (-x) + (-x)^2} - \sqrt{1 - (-x) + (-x)^2}. \]

This simplifies to:

\[ f(-x) = \sqrt{1 - x + x^2} - \sqrt{1 + x + x^2}. \]

### Step 3: Rearranging \( f(-x) \)
Now, we can rearrange \( f(-x) \):

\[ f(-x) = -\left( \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2} \right) = -f(x). \]

### Step 4: Conclusion about odd function
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 5: Apply the property of definite integrals
For an odd function integrated over a symmetric interval about zero, the integral evaluates to zero:

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

Thus, we have:

\[ I = \int_{-1}^{1} f(x) \, dx = 0. \]

### Final Answer
Therefore, the value of the definite integral is

\[ \boxed{0}. \]

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