If `|x| lt 1` then coefficient of `x^(2)` in `(log(1+x))/((1-x)^(2))` is

Options

A. `(3)/(2)`
B. 2
C. `-(1)/(2)`
D. 4

Answer

To find the coefficient of \( x^2 \) in the expression \( \frac{\log(1+x)}{(1-x)^2} \) for \( |x| < 1 \), we can follow these steps:

### Step 1: Expand \( \log(1+x) \)
The Taylor series expansion of \( \log(1+x) \) around \( x = 0 \) is given by:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]
This series converges for \( |x| < 1 \).

### Step 2: Expand \( (1-x)^{-2} \)
Using the binomial series, we can expand \( (1-x)^{-2} \):
\[ (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \cdots \]
This series also converges for \( |x| < 1 \).

### Step 3: Multiply the two expansions
Now we need to multiply the two series obtained in Steps 1 and 2:
\[ \log(1+x) \cdot (1-x)^{-2} = \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \right) \cdot \left( 1 + 2x + 3x^2 + 4x^3 + \cdots \right) \]

### Step 4: Collect the terms for \( x^2 \)
To find the coefficient of \( x^2 \), we consider the contributions from the products of terms that yield \( x^2 \):
1. From \( x \) in \( \log(1+x) \) and \( 2x \) in \( (1-x)^{-2} \):
\[ x \cdot 2x = 2x^2 \]
2. From \( -\frac{x^2}{2} \) in \( \log(1+x) \) and the constant term \( 1 \) in \( (1-x)^{-2} \):
\[ -\frac{x^2}{2} \cdot 1 = -\frac{x^2}{2} \]

Adding these contributions together gives:
\[ 2x^2 - \frac{x^2}{2} = \left( 2 - \frac{1}{2} \right)x^2 = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}x^2 \]

### Conclusion
Thus, the coefficient of \( x^2 \) in the expression \( \frac{\log(1+x)}{(1-x)^2} \) is:
\[ \frac{3}{2} \]