The coefficient of `x^(6)` in
`log[(1+x)^(1+x)(1-x)^(1-x)]=`

Options

A. a) `(1)/(15)`
B. b) `(1)/(30)`
C. c) `(1)/(10)`
D. d) `(1)/(45)`

Answer

To find the coefficient of \( x^6 \) in the expression

\[ \log\left[(1+x)^{1+x}(1-x)^{1-x}\right], \]

we can follow these steps:

### Step 1: Use the logarithmic property
Using the property of logarithms that states \( \log(A \cdot B) = \log A + \log B \), we can rewrite the expression as:

\[ \log\left[(1+x)^{1+x}\right] + \log\left[(1-x)^{1-x}\right]. \]

### Step 2: Simplify using logarithmic identities
Applying the power rule of logarithms, we have:

\[ (1+x) \log(1+x) + (1-x) \log(1-x). \]

### Step 3: Expand the logarithmic functions
We can expand \( \log(1+x) \) and \( \log(1-x) \) using their Taylor series expansions:

\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]

\[ \log(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots\right). \]

### Step 4: Substitute the expansions
Substituting these expansions into our expression gives:

\[ (1+x)\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\right) + (1-x)\left(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots\right). \]

### Step 5: Distribute and collect like terms
Now we need to distribute and collect terms for \( x^6 \):

1. From \( (1+x)\log(1+x) \):
- The term \( 1 \cdot x^6 \) contributes \( \frac{1}{6} \).
- The term \( x \cdot x^5 \) contributes \( \frac{1}{5} \).

2. From \( (1-x)\log(1-x) \):
- The term \( 1 \cdot (-x^6) \) contributes \( -\frac{1}{6} \).
- The term \( -x \cdot (-x^5) \) contributes \( \frac{1}{5} \).

### Step 6: Combine contributions
Now we combine all contributions to find the coefficient of \( x^6 \):

\[ \text{Coefficient of } x^6 = \left(\frac{1}{6} - \frac{1}{6}\right) + \left(\frac{1}{5} + \frac{1}{5}\right) = 0 + \frac{2}{5} = \frac{2}{5}. \]

### Step 7: Final calculation
Now we calculate the final coefficient:

\[ \frac{2}{5} - \frac{2}{6} = \frac{2}{5} - \frac{5}{15} = \frac{6}{15} - \frac{5}{15} = \frac{1}{15}. \]

Thus, the coefficient of \( x^6 \) in the expansion is

\[ \boxed{\frac{1}{15}}. \]