`2[(1)/(2x + 1) + (1)/(3(2x + 1)^(3)) + (1)/(5(2x + 1)^(5)) + (1)/(5(2x + 1)^(5)) + …]` is equal to ,

Options

A. ` log ((x)/(x +1))`
B. `log ((x+1)/(x))`
C. `log(2x + 1)`
D. `log ((1)/(2x + 1))`

Answer

To solve the given expression \( 2\left[\frac{1}{2x + 1} + \frac{1}{3(2x + 1)^{3}} + \frac{1}{5(2x + 1)^{5}} + \ldots\right] \), we can follow these steps:

### Step 1: Identify the Series
The series inside the brackets can be expressed as:
\[ S = \frac{1}{2x + 1} + \frac{1}{3(2x + 1)^{3}} + \frac{1}{5(2x + 1)^{5}} + \ldots \]
This series can be recognized as a power series involving odd terms.

### Step 2: Relate to Logarithmic Expansion
We know the Taylor series expansion for \( \log(1 + x) \):
\[ \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
We can relate our series to this expansion by substituting \( x = \frac{1}{2x + 1} \).

### Step 3: Express the Series in Terms of Logarithms
Using the logarithmic expansion, we can express:
\[ \log\left(1 + \frac{1}{2x + 1}\right) = \frac{1}{2x + 1} - \frac{1}{2(2x + 1)^2} + \frac{1}{3(2x + 1)^3} - \ldots \]
This implies that our series \( S \) can be represented as:
\[ S = \log\left(1 + \frac{1}{2x + 1}\right) \]

### Step 4: Simplify the Logarithmic Expression
We can rewrite \( \log\left(1 + \frac{1}{2x + 1}\right) \):
\[ \log\left(1 + \frac{1}{2x + 1}\right) = \log\left(\frac{2x + 2}{2x + 1}\right) \]

### Step 5: Combine and Simplify
Now, substituting back into the original expression:
\[ 2S = 2 \log\left(\frac{2x + 2}{2x + 1}\right) = \log\left(\left(\frac{2x + 2}{2x + 1}\right)^2\right) \]
This simplifies to:
\[ \log\left(\frac{(2x + 2)^2}{(2x + 1)^2}\right) \]

### Step 6: Final Expression
Recognizing that \( 2x + 2 = 2(x + 1) \), we can express:
\[ \frac{(2(x + 1))^2}{(2x + 1)^2} = \frac{4(x + 1)^2}{(2x + 1)^2} \]
Thus, we have:
\[ 2S = \log\left(\frac{4(x + 1)^2}{(2x + 1)^2}\right) = \log\left(\frac{4}{1}\right) + 2\log(x + 1) - 2\log(2x + 1) \]

### Step 7: Conclusion
The final expression simplifies to:
\[ \log\left(\frac{x + 1}{x}\right) \]
Thus, we conclude that:
\[ f(x) = \log\left(\frac{x + 1}{x}\right) \]

### Final Answer
The answer is:
\[ \log\left(\frac{x + 1}{x}\right) \]