If higher powers of `x^(2)` are neglected, then the value of `log(1+x^(2))-log(1+x)-log(1-x)=`

Options

A. `x^(2)`
B. `(1)/(2)x^(2)`
C. `2x^(2)`
D. `3x^(2)`

Answer

To solve the problem \( \log(1+x^2) - \log(1+x) - \log(1-x) \) while neglecting higher powers of \( x^2 \), we can use the Taylor series expansion for logarithmic functions.

### Step-by-Step Solution:

1. **Use the logarithmic identity**:
\[ \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \]
Therefore, we can rewrite the expression:
\[ \log(1+x^2) - \log(1+x) - \log(1-x) = \log\left(\frac{1+x^2}{(1+x)(1-x)}\right) \]

2. **Simplify the denominator**:
The denominator can be simplified using the difference of squares:
\[ (1+x)(1-x) = 1 - x^2 \]
Thus, we have:
\[ \log\left(\frac{1+x^2}{1-x^2}\right) \]

3. **Expand using Taylor series**:
We can use the Taylor series expansion for \( \log(1+y) \):
\[ \log(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \cdots \]
Applying this to \( \log(1+x^2) \) and \( \log(1-x^2) \):
- For \( \log(1+x^2) \):
\[ \log(1+x^2) \approx x^2 - \frac{x^4}{2} \quad \text{(neglecting higher powers)} \]
- For \( \log(1-x^2) \):
\[ \log(1-x^2) \approx -x^2 - \frac{(-x^2)^2}{2} = -x^2 + \frac{x^4}{2} \quad \text{(neglecting higher powers)} \]

4. **Combine the expansions**:
Now, substituting these expansions back into our expression:
\[ \log(1+x^2) - \log(1-x^2) \approx \left(x^2 - \frac{x^4}{2}\right) - \left(-x^2 + \frac{x^4}{2}\right) \]
Simplifying this gives:
\[ x^2 - \frac{x^4}{2} + x^2 - \frac{x^4}{2} = 2x^2 - x^4 \]

5. **Neglect higher powers**:
Since we are neglecting higher powers of \( x^2 \), we ignore the \( -x^4 \) term:
\[ \approx 2x^2 \]

### Final Answer:
Thus, the value of \( \log(1+x^2) - \log(1+x) - \log(1-x) \) is:
\[ \boxed{2x^2} \]