The larger of log (1 + x) and `( tan^(-1) x)/(1 +x)` when x>0 is ..........

Answer

To determine which of the two functions, \( \log(1 + x) \) or \( \frac{\tan^{-1}(x)}{1 + x} \), is larger for \( x > 0 \), we will analyze the difference between these two functions.

### Step-by-Step Solution

1. **Define the Functions**:
Let:
\[ f(x) = \log(1 + x) - \frac{\tan^{-1}(x)}{1 + x} \]
We want to determine the sign of \( f(x) \) for \( x > 0 \).

2. **Calculate the Derivative**:
To analyze \( f(x) \), we compute its derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx} \left( \log(1 + x) \right) - \frac{d}{dx} \left( \frac{\tan^{-1}(x)}{1 + x} \right) \]
The derivative of \( \log(1 + x) \) is:
\[ \frac{1}{1 + x} \]
For the second term, we apply the quotient rule:
\[ \frac{d}{dx} \left( \frac{\tan^{-1}(x)}{1 + x} \right) = \frac{(1 + x) \cdot \frac{1}{1 + x^2} - \tan^{-1}(x)}{(1 + x)^2} \]
Therefore, we have:
\[ f'(x) = \frac{1}{1 + x} - \frac{(1 + x) \cdot \frac{1}{1 + x^2} - \tan^{-1}(x)}{(1 + x)^2} \]

3. **Simplify the Derivative**:
Combine the terms:
\[ f'(x) = \frac{1}{1 + x} - \frac{1}{1 + x^2} + \frac{\tan^{-1}(x)}{(1 + x)^2} \]
Now, we need to analyze the sign of \( f'(x) \).

4. **Check the Sign of \( f'(x) \)**:
For \( x > 0 \):
- \( \frac{1}{1 + x} > 0 \)
- \( \frac{1}{1 + x^2} > 0 \)
- \( \tan^{-1}(x) > 0 \)

Since all terms are positive, we can conclude that \( f'(x) > 0 \) for \( x > 0 \). This means \( f(x) \) is an increasing function.

5. **Evaluate \( f(0) \)**:
Now, we evaluate \( f(0) \):
\[ f(0) = \log(1 + 0) - \frac{\tan^{-1}(0)}{1 + 0} = 0 - 0 = 0 \]

6. **Conclusion**:
Since \( f(0) = 0 \) and \( f(x) \) is increasing for \( x > 0 \), it follows that \( f(x) > 0 \) for all \( x > 0 \). Therefore:
\[ \log(1 + x) > \frac{\tan^{-1}(x)}{1 + x} \text{ for all } x > 0 \]

### Final Answer:
The larger of \( \log(1 + x) \) and \( \frac{\tan^{-1}(x)}{1 + x} \) when \( x > 0 \) is \( \log(1 + x) \).