`int _(-9) ^(9) (x ^(3))/( 4- x ^(2)) dx =`

Options

A. 0
B. 3
C. 9
D. `-9`

Answer

To solve the integral

\[ I = \int_{-9}^{9} \frac{x^3}{4 - x^2} \, dx, \]

we can use the property of definite integrals concerning odd and even functions.

### Step 1: Identify the function
Let

\[ f(x) = \frac{x^3}{4 - x^2}. \]

### Step 2: Check if the function is odd or even
To determine if \( f(x) \) is odd or even, we need to evaluate \( f(-x) \):

\[ f(-x) = \frac{(-x)^3}{4 - (-x)^2} = \frac{-x^3}{4 - x^2}. \]

### Step 3: Compare \( f(-x) \) with \( -f(x) \)
Now, we can see that:

\[ -f(x) = -\left(\frac{x^3}{4 - x^2}\right) = \frac{-x^3}{4 - x^2}. \]

Thus, we have:

\[ f(-x) = -f(x). \]

This means that \( f(x) \) is an odd function.

### Step 4: Use the property of odd functions in integrals
Since \( f(x) \) is an odd function, we can apply the property of integrals:

\[ \int_{-a}^{a} f(x) \, dx = 0 \quad \text{if } f(x) \text{ is odd}. \]

### Step 5: Conclude the value of the integral
Therefore, we conclude that:

\[ I = \int_{-9}^{9} \frac{x^3}{4 - x^2} \, dx = 0. \]

### Final Answer:
\[ \int_{-9}^{9} \frac{x^3}{4 - x^2} \, dx = 0. \]

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