The function $f(x) = \log(1 + x) - \frac{2x}{2 + x}$ is increasing on

Domain first: $1+x>0$, so $x>-1$, and also $x\neq -2$, which is already outside the domain. Thus the domain is $(-1,\infty)$. Differentiate: $$f'(x)=\frac{1}{1+x}-\frac{4}{(2+x)^2}.$$ Take the LCM and simplify: $$f'(x)=\frac{(x+2)^2-4(x+1)}{(x+1)(x+2)^2}.$$ $$f'(x)=\frac{x^2}{(x+1)(x+2)^2}.$$ For every $x>-1$, we have $(x+1)>0$ and $(x+2)^2>0$, while $x^2\ge 0$. Hence $f'(x)\ge 0$ on the whole domain, with equality only at $x=0$. Therefore $f(x)$ is increasing on $(-1,\infty)$. Now compare with the options: $(0,\infty)$ is true but not the full interval; $(-\infty,0)$ and $(-\infty,\infty)$ are impossible because the function is not even defined there. Since $(-1,\infty)$ is not listed, the correct choice is $\text{None of these}$.