Find which of the two larger `log(1+x) " or " (tan^(-1)x)/(1+x).`

Answer

To determine which of the two functions \( \log(1+x) \) or \( \frac{\tan^{-1}(x)}{1+x} \) is larger, we can analyze their behavior and ranges. Here’s a step-by-step solution:

### Step 1: Define the Functions
Let:
- \( f(x) = \log(1+x) \)
- \( g(x) = \frac{\tan^{-1}(x)}{1+x} \)

### Step 2: Determine the Domain
Both functions are defined for \( x > -1 \) (since \( \log(1+x) \) is only defined for \( x > -1 \)). The function \( \tan^{-1}(x) \) is defined for all real numbers.

### Step 3: Analyze the Range of \( f(x) \)
The function \( f(x) = \log(1+x) \):
- As \( x \to -1^+ \), \( f(x) \to -\infty \).
- As \( x \to \infty \), \( f(x) \to \infty \).
- Therefore, the range of \( f(x) \) is \( (-\infty, \infty) \).

### Step 4: Analyze the Range of \( g(x) \)
The function \( g(x) = \frac{\tan^{-1}(x)}{1+x} \):
- The range of \( \tan^{-1}(x) \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
- As \( x \to -1^+ \), \( g(x) \to \frac{\tan^{-1}(-1)}{0} \) which is undefined.
- As \( x \to 0 \), \( g(0) = \frac{\tan^{-1}(0)}{1} = 0 \).
- As \( x \to \infty \), \( g(x) \to \frac{\frac{\pi}{2}}{1+\infty} = 0 \).
- Therefore, \( g(x) \) approaches 0 but does not reach it, and it remains positive for \( x > 0 \).

### Step 5: Compare the Two Functions
- Since \( f(x) \) has a range of \( (-\infty, \infty) \) and \( g(x) \) approaches 0 but does not reach it, we can conclude that for sufficiently large \( x \), \( f(x) \) will always be larger than \( g(x) \).
- Additionally, for \( x = 0 \), \( f(0) = \log(1) = 0 \) and \( g(0) = 0 \). However, for \( x > 0 \), \( f(x) \) increases without bound while \( g(x) \) remains bounded and approaches 0.

### Conclusion
Thus, we conclude that:
\[ \log(1+x) > \frac{\tan^{-1}(x)}{1+x} \quad \text{for all } x > 0. \]