`int_(-1)^(1) (sin x-x^(2))/(3-|x|)=`

Options

A. 0
B. `2 int_(1)^(0) (sin x)/(3-|x|) dx`
C. `2 int_(0)^(1) (-x^(2))/(3-|x|) dx`
D. `2 int_(0)^(1) (sin x-x^(2))/(3-|x|) -dx`

Answer

To solve the integral

\[ I = \int_{-1}^{1} \frac{\sin x - x^2}{3 - |x|} \, dx, \]

we can split this integral into two parts based on the properties of the functions involved.

### Step 1: Split the Integral
We can write the integral as:

\[ I = \int_{-1}^{1} \frac{\sin x}{3 - |x|} \, dx - \int_{-1}^{1} \frac{x^2}{3 - |x|} \, dx. \]

Let’s denote these two integrals as \( I_1 \) and \( I_2 \):

\[ I_1 = \int_{-1}^{1} \frac{\sin x}{3 - |x|} \, dx, \]
\[ I_2 = \int_{-1}^{1} \frac{x^2}{3 - |x|} \, dx. \]

### Step 2: Analyze \( I_1 \)
To determine if \( I_1 \) is zero, we check if the function \( f(x) = \frac{\sin x}{3 - |x|} \) is odd.

Calculating \( f(-x) \):

\[ f(-x) = \frac{\sin(-x)}{3 - |-x|} = \frac{-\sin x}{3 - |x|} = -f(x). \]

Since \( f(-x) = -f(x) \), \( f(x) \) is an odd function. Therefore, the integral of an odd function over a symmetric interval around zero is zero:

\[ I_1 = 0. \]

### Step 3: Analyze \( I_2 \)
Now, we check \( I_2 \):

\[ I_2 = \int_{-1}^{1} \frac{x^2}{3 - |x|} \, dx. \]

To check if this function is even, we compute \( f(-x) \):

\[ f(-x) = \frac{(-x)^2}{3 - |-x|} = \frac{x^2}{3 - |x|} = f(x). \]

Since \( f(-x) = f(x) \), \( f(x) \) is an even function. Therefore, we can use the property of even functions:

\[ I_2 = 2 \int_{0}^{1} \frac{x^2}{3 - x} \, dx. \]

### Step 4: Substitute Back
Now we can substitute back into our original integral:

\[ I = I_1 - I_2 = 0 - 2 \int_{0}^{1} \frac{x^2}{3 - x} \, dx = -2 \int_{0}^{1} \frac{x^2}{3 - x} \, dx. \]

### Step 5: Evaluate the Integral
Now we need to evaluate \( \int_{0}^{1} \frac{x^2}{3 - x} \, dx \). We can use integration by parts or a substitution method to solve this integral.

Let’s perform the integration:

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

Using the substitution \( u = 3 - x \), \( du = -dx \), we can change the limits accordingly:

When \( x = 0 \), \( u = 3 \) and when \( x = 1 \), \( u = 2 \).

Thus,

\[ \int_{0}^{1} \frac{x^2}{3 - x} \, dx = \int_{3}^{2} \frac{(3 - u)^2}{u} (-du) = \int_{2}^{3} \frac{(3 - u)^2}{u} \, du. \]

Now we can evaluate this integral.

### Final Result
After evaluating the integral, we can substitute back to find \( I \):

\[ I = -2 \cdot \text{(value of the integral)}. \]