The function f(x) = log (1 + x) – 2x/2+x is increasing on

The correct option is (-1, ∞).

To determine where the function is increasing, we first note the domain of the function.
The function is given by:
$$f(x)=\log (1+x)-\frac{2x}{2+x}$$

For the logarithmic term $\log (1+x)$ to be defined, we must have:
$$1+x>0\Rightarrow x>-1$$ Thus, the domain of the function is $(-1,\infty )$.

A function $f(x)$ is increasing on an interval if its first derivative is positive, i.e., $f^{\prime }(x)>0$ on that interval.
Let us find the derivative $f^{\prime }(x)$ with respect to $x$: $$f^{\prime }(x)=\frac{d}{dx}[\log (1+x)]-\frac{d}{dx}\left[\frac{2x}{2+x}\right]$$

Using the standard derivative rule for log and the quotient rule for the fraction:
$$f^{\prime }(x)=\frac{1}{1+x}-\frac{(2+x)·2-2x·1}{{(2+x)}^2}$$

Simplifying the numerator of the second term:
$$f^{\prime }(x)=\frac{1}{1+x}-\frac{4+2x-2x}{{(2+x)}^2}$$
$$f^{\prime }(x)=\frac{1}{1+x}-\frac{4}{{(2+x)}^2}$$

Now, let's combine these terms under a common denominator:
$$f^{\prime }(x)=\frac{{(2+x)}^2-4(1+x)}{(1+x){(2+x)}^2}$$
$$f^{\prime }(x)=\frac{(4+4x+x^2)-(4+4x)}{(1+x){(2+x)}^2}$$
$$f^{\prime }(x)=\frac{x^2}{(1+x){(2+x)}^2}$$

To analyze the sign of $f^{\prime }(x)$ in the domain $x>-1$:
1. The numerator $x^2\ge 0$ for all real values of $x$.
2. The term ${(2+x)}^2>0$ for all $x\ne -2$.
3. The term $(1+x)>0$ because the domain restricts $x>-1$.

Thus, for all $x>-1$ (excluding $x=0$ where the derivative is zero):
$$f^{\prime }(x)>0$$ Since $f^{\prime }(x)\ge 0$ for all $x\in (-1,\infty )$ and equality holds only at a single point ($x=0$), the function is strictly increasing on the entire domain $(-1,\infty )$.