The function `x-(log(1+x)/x)(x>0)` is increasing in:

Options

A. `(1,oo)`
B. `(0,oo)`
C. `(2,2e)`
D. `(1//e,2e)`

Answer

To determine the intervals where the function \( f(x) = x - \frac{\log(1+x)}{x} \) is increasing for \( x > 0 \), we need to find the derivative \( f'(x) \) and analyze its sign.

### Step 1: Find the derivative \( f'(x) \)

Given:
\[ f(x) = x - \frac{\log(1+x)}{x} \]

We can differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(x\right) - \frac{d}{dx}\left(\frac{\log(1+x)}{x}\right) \]

The derivative of \( x \) is \( 1 \). For the second term, we will use the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
where \( u = \log(1+x) \) and \( v = x \).

Calculating \( u' \) and \( v' \):
- \( u' = \frac{1}{1+x} \)
- \( v' = 1 \)

Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{\log(1+x)}{x}\right) = \frac{\left(\frac{1}{1+x}\right)x - \log(1+x)(1)}{x^2} \]
\[ = \frac{x}{(1+x)x} - \frac{\log(1+x)}{x^2} \]
\[ = \frac{1}{1+x} - \frac{\log(1+x)}{x^2} \]

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

### Step 2: Analyze the sign of \( f'(x) \)

To determine where \( f(x) \) is increasing, we need \( f'(x) > 0 \):
\[ \frac{x}{1+x} + \frac{\log(1+x)}{x^2} > 0 \]

Since \( x > 0 \), we know:
- \( \frac{x}{1+x} > 0 \)
- \( \log(1+x) > 0 \) for \( x > 0 \)

Thus, both terms are positive for \( x > 0 \).

### Step 3: Conclusion

Since \( f'(x) > 0 \) for all \( x > 0 \), the function \( f(x) \) is increasing for \( x > 0 \).

### Final Answer

The function \( f(x) \) is increasing in the interval \( (0, \infty) \).

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