The value of `int_(-1)^(1) {sqrt(1+x+x^(2))-sqrt(1-x+x^(2))}dx`, is

Options

A. 0
B. 1
C. -1
D. none of these

Explanation

To solve the integral

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

we will follow these steps:

### Step 1: Define the Function
Let

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

### Step 2: Check for Odd Function Property
To check if \( f(x) \) is an odd function, we need to evaluate \( f(-x) \):

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

### Step 3: Simplify \( f(-x) \)
Rearranging gives:

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

### Step 4: Conclusion on 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 odd functions, the property states that:

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

Thus, we can apply this property to our integral:

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

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

\[ \boxed{0}. \]