Find the interval in which

we have

$f(x)=\log (1+x)-\frac{x}{1+x}$

$f^{\prime}(x)=\frac{1}{1+x}-\left(\frac{(1+x)-x}{(1+x)^{2}}\right)$

$=\frac{1}{1+x}-\left(\frac{1}{(1+x)^{2}}\right)$

$=\frac{x}{(1+x)^{2}}$

Critical points

$\mathrm{f}^{\prime}(\mathrm{x})=0$

$\Rightarrow \frac{x}{(1+x)^{2}}=0$

$\Rightarrow x=0,-1$

Clearly, $f^{\prime}(x)>0$ if $x>0$

And $f^{\prime}(x)<0$ if $-1

Hence, $f(x)$ increases in $(0, \infty)$, decreases in $(-\infty,-1) \cup(-1,0)$