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

Given, \(f(x)=\log (1+x)-\frac{2 x}{2+x}\)
Differentiating the function w.r.t. \(x\), we get
\[
\begin{aligned}
f^{\prime}(x) & =\frac{1}{(1+x)}(0+1)-2\left[\frac{(2+x) \times 1-x(0+1)}{(2+x)^2}\right] \\
& =\frac{1}{1+x}-2\left[\frac{2+x-x}{(2+x)^2}\right]=\frac{1}{1+x}-\frac{4}{(2+x)^2} \\
& =\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2}=\frac{4+x^2+4 x-4-4 x}{(x+1)(x+2)^2} \\
& =\frac{x^2}{(x+1)(x+2)^2}=\left(\frac{x}{x+2}\right)^2 \cdot \frac{1}{(x+1)}
\end{aligned}
\]
Sign of \(f^{\prime}(x)\) depends on the sign of \(\frac{1}{(x+1)}\)
\[
\begin{aligned}
& f^{\prime}(x)>0 \text { when } x>-1 \Rightarrow \text { increasing } \\
& f^{\prime}(x) < 0 \text { when } x < -1 \Rightarrow \text { decreasing }
\end{aligned}
\]
Hence, \(f(x)\) is increasing on \((-1, \infty)\).