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

Options

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

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 follow these steps:

### Step 1: Define the integral
We start by defining the integral:

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

### Step 2: Use the property of definite integrals
We can utilize the property of definite integrals that states:

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

In our case, \( a = -1 \) and \( b = 1 \), thus \( a + b = 0 \). We will replace \( x \) with \( -x \):

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

### Step 3: Simplify the expression
Now we simplify the terms inside the integral:

1. For \( \sqrt{1+(-x)+(-x)^2} \):
\[ \sqrt{1-x+x^2}. \]

2. For \( \sqrt{1-(-x)+(-x)^2} \):
\[ \sqrt{1+x+x^2}. \]

Thus, we can rewrite \( I \):

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

### Step 4: Combine the two integrals
Now we have two expressions for \( I \):

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

Adding these two equations gives:

\[ 2I = \int_{-1}^{1} \left( \sqrt{1+x+x^2} - \sqrt{1+x+x^2} \right) dx = 0. \]

### Step 5: Solve for \( I \)
Since \( 2I = 0 \), we can conclude that:

\[ I = 0. \]

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

\[ \boxed{0}. \]