The function `log(1+x) - (2x)/(x+2)` is increasing in the interval:

Options

A. `(-infty, 0]`
B. `(-1, infty)`
C. `(-infty, 1]`
D. None of these

Answer

To determine the interval in which the function \( f(x) = \log(1+x) - \frac{2x}{x+2} \) is increasing, we need to find the derivative of the function and analyze where it is greater than zero.

### Step 1: Find the derivative of the function
The function is given as:
\[ f(x) = \log(1+x) - \frac{2x}{x+2} \]

To find \( f'(x) \), we differentiate each term separately.

1. The derivative of \( \log(1+x) \) is:
\[ \frac{d}{dx} \log(1+x) = \frac{1}{1+x} \]

2. For the second term \( -\frac{2x}{x+2} \), we use the quotient rule. If \( u = 2x \) and \( v = x+2 \), then:
\[ \frac{d}{dx} \left(-\frac{u}{v}\right) = -\frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( \frac{du}{dx} = 2 \) and \( \frac{dv}{dx} = 1 \).

Applying the quotient rule:
\[ -\frac{(x+2)(2) - (2x)(1)}{(x+2)^2} = -\frac{2x + 4 - 2x}{(x+2)^2} = -\frac{4}{(x+2)^2} \]

Combining these results, we have:
\[ f'(x) = \frac{1}{1+x} - \frac{4}{(x+2)^2} \]

### Step 2: Set the derivative greater than zero
To find where the function is increasing, we set the derivative greater than zero:
\[ \frac{1}{1+x} - \frac{4}{(x+2)^2} > 0 \]

### Step 3: Solve the inequality
Rearranging gives:
\[ \frac{1}{1+x} > \frac{4}{(x+2)^2} \]

Cross-multiplying (noting that both sides are positive for \( x > -1 \)):
\[ (x+2)^2 > 4(1+x) \]

Expanding both sides:
\[ x^2 + 4x + 4 > 4 + 4x \]
\[ x^2 > 0 \]

### Step 4: Analyze the inequality
The inequality \( x^2 > 0 \) holds for all \( x \neq 0 \). However, we must also consider the domain of the original function. The function \( \log(1+x) \) is defined for \( x > -1 \).

### Step 5: Combine the results
Thus, the function is increasing for:
\[ x > 0 \]

### Conclusion
The function \( f(x) = \log(1+x) - \frac{2x}{x+2} \) is increasing in the interval:
\[ (0, \infty) \]