`int_(-1)^(1) (1+x^(3))/(9-x^(2)) dx =`

Options

A. `2/3 log 2`
B. `1/6 log 2`
C. `1/3 log 2`
D. `3/2 log 2`

Answer

To solve the integral \( I = \int_{-1}^{1} \frac{1 + x^3}{9 - x^2} \, dx \), we can use the property of definite integrals that states:

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

### Step 1: Evaluate \( I \) using the property of symmetry

First, we compute \( I \) by substituting \( -x \) into the integrand:

\[ I = \int_{-1}^{1} \frac{1 + (-x)^3}{9 - (-x)^2} \, dx = \int_{-1}^{1} \frac{1 - x^3}{9 - x^2} \, dx \]

### Step 2: Add the two integrals

Now we have two expressions for \( I \):

1. \( I = \int_{-1}^{1} \frac{1 + x^3}{9 - x^2} \, dx \)
2. \( I = \int_{-1}^{1} \frac{1 - x^3}{9 - x^2} \, dx \)

Adding these two equations gives:

\[ 2I = \int_{-1}^{1} \left( \frac{1 + x^3}{9 - x^2} + \frac{1 - x^3}{9 - x^2} \right) \, dx \]

### Step 3: Simplify the integrand

The integrand simplifies as follows:

\[ \frac{1 + x^3 + 1 - x^3}{9 - x^2} = \frac{2}{9 - x^2} \]

Thus, we have:

\[ 2I = \int_{-1}^{1} \frac{2}{9 - x^2} \, dx \]

### Step 4: Factor out the constant

We can factor out the constant \( 2 \):

\[ 2I = 2 \int_{-1}^{1} \frac{1}{9 - x^2} \, dx \]

Dividing both sides by 2 gives:

\[ I = \int_{-1}^{1} \frac{1}{9 - x^2} \, dx \]

### Step 5: Solve the integral

To solve \( \int_{-1}^{1} \frac{1}{9 - x^2} \, dx \), we can use the formula for the integral of the form \( \int \frac{1}{a^2 - x^2} \, dx \):

\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C \]

In our case, \( a^2 = 9 \) so \( a = 3 \). Thus, we have:

\[ \int \frac{1}{9 - x^2} \, dx = \frac{1}{6} \ln \left| \frac{3 + x}{3 - x} \right| + C \]

### Step 6: Evaluate the definite integral

Now we evaluate from \( -1 \) to \( 1 \):

\[ I = \left[ \frac{1}{6} \ln \left| \frac{3 + x}{3 - x} \right| \right]_{-1}^{1} \]

Calculating at the limits:

1. At \( x = 1 \):

\[ \frac{1}{6} \ln \left| \frac{3 + 1}{3 - 1} \right| = \frac{1}{6} \ln \left| \frac{4}{2} \right| = \frac{1}{6} \ln 2 \]

2. At \( x = -1 \):

\[ \frac{1}{6} \ln \left| \frac{3 - 1}{3 + 1} \right| = \frac{1}{6} \ln \left| \frac{2}{4} \right| = \frac{1}{6} \ln \frac{1}{2} = -\frac{1}{6} \ln 2 \]

### Step 7: Combine the results

Now we combine the results:

\[ I = \frac{1}{6} \ln 2 - \left(-\frac{1}{6} \ln 2\right) = \frac{1}{6} \ln 2 + \frac{1}{6} \ln 2 = \frac{2}{6} \ln 2 = \frac{1}{3} \ln 2 \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{-1}^{1} \frac{1 + x^3}{9 - x^2} \, dx = \frac{1}{3} \ln 2 \]

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