The sum of the series
`(x-1)/(x+1)+1/2(x^(2)-1)/(x+1)^(2)+1/3(x^(3)-1)/(x+1)^(3)`+…is equal to

Options

A. `log_(e)x`
B. `2 log_(e)x`
C. `-log_(e)(x+1)`
D. none of these

Answer

To find the sum of the series

\[ S = \frac{x-1}{x+1} + \frac{1}{2} \cdot \frac{x^2-1}{(x+1)^2} + \frac{1}{3} \cdot \frac{x^3-1}{(x+1)^3} + \ldots \]

we can break down the series step by step.

### Step 1: Rewrite Each Term
Each term in the series can be rewritten. The first term is:

\[ \frac{x-1}{x+1} = \frac{x}{x+1} - \frac{1}{x+1} \]

Similarly, the second term can be rewritten as:

\[ \frac{1}{2} \cdot \frac{x^2-1}{(x+1)^2} = \frac{x^2}{2(x+1)^2} - \frac{1}{2(x+1)^2} \]

And the third term is:

\[ \frac{1}{3} \cdot \frac{x^3-1}{(x+1)^3} = \frac{x^3}{3(x+1)^3} - \frac{1}{3(x+1)^3} \]

So, we can express the series as:

\[ S = \left( \frac{x}{x+1} + \frac{x^2}{2(x+1)^2} + \frac{x^3}{3(x+1)^3} + \ldots \right) - \left( \frac{1}{x+1} + \frac{1}{2(x+1)^2} + \frac{1}{3(x+1)^3} + \ldots \right) \]

### Step 2: Identify the Two Series
Let’s denote the two parts of the series:

1. **Positive Part**:
\[ S_1 = \frac{x}{x+1} + \frac{x^2}{2(x+1)^2} + \frac{x^3}{3(x+1)^3} + \ldots \]

2. **Negative Part**:
\[ S_2 = \frac{1}{x+1} + \frac{1}{2(x+1)^2} + \frac{1}{3(x+1)^3} + \ldots \]

### Step 3: Recognize the Series as Logarithmic Forms
The positive part \( S_1 \) can be recognized as a series that resembles the Taylor series expansion for \(-\log(1-x)\):

\[ S_1 = -\log\left(1 - \frac{x}{x+1}\right) = -\log\left(\frac{1}{x+1}\right) = \log(x+1) \]

The negative part \( S_2 \) can be expressed similarly:

\[ S_2 = -\log\left(1 - \frac{1}{x+1}\right) = -\log\left(\frac{x}{x+1}\right) = \log(x+1) - \log(x) \]

### Step 4: Combine the Results
Now, substituting back into the expression for \( S \):

\[ S = S_1 - S_2 = \log(x+1) - \left(\log(x+1) - \log(x)\right) \]

This simplifies to:

\[ S = \log(x) \]

### Final Answer
Thus, the sum of the series is:

\[ \boxed{\log(x)} \]