`2((2x)/(x^(2)+1)+(1)/(3)((2x)/(x^(2)+1))^(3) +(1)/(5)((2x)/(x^(2)+1))^(5)+…..oo)=`

Options

A. a) `log_(e )((x+1)/(x-1))`
B. b)`log_(e )((x-1)/(x+1))`
C. c) `log_(e )((1)/(x-1))`
D. d) `log_(e )((1)/(x+1))`

Answer

To solve the problem, we need to evaluate the series:

\[ 2\left(\frac{2x}{x^2 + 1} + \frac{1}{3}\left(\frac{2x}{x^2 + 1}\right)^3 + \frac{1}{5}\left(\frac{2x}{x^2 + 1}\right)^5 + \ldots\right) \]

### Step 1: Recognize the Series Structure
The series can be recognized as a Taylor series expansion of logarithmic functions. Specifically, we can relate it to the series for \(\log(1+x)\) and \(\log(1-x)\).

### Step 2: Write Down the Logarithmic Series
The Taylor series expansions for \(\log(1+x)\) and \(\log(1-x)\) are:

\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \ldots \]

\[ \log(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + \ldots\right) \]

### Step 3: Combine the Series
To find a series that includes only odd powers, we can subtract the two series:

\[ \log(1+x) - \log(1-x) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]

This results in:

\[ \log\left(\frac{1+x}{1-x}\right) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]

### Step 4: Substitute \(x\) in the Series
In our case, we can set \(x = \frac{2x}{x^2 + 1}\). Thus, we need to evaluate:

\[ \log\left(\frac{1 + \frac{2x}{x^2 + 1}}{1 - \frac{2x}{x^2 + 1}}\right) \]

### Step 5: Simplify the Expression
To simplify the expression, we need to find a common denominator:

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

### Step 6: Apply the Logarithm
Now, we can apply the logarithm:

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

### Step 7: Multiply by 2
Finally, we multiply the entire expression by 2:

\[ 2 \cdot 2\log\left(\frac{x + 1}{x - 1}\right) = 4\log\left(\frac{x + 1}{x - 1}\right) \]

### Final Answer
Thus, the final answer is:

\[ \boxed{4\log\left(\frac{x + 1}{x - 1}\right)} \]