Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$ Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$
One method is considering a function $f(x) = x − \ln(1 + x) − \frac{x^2}{2}$ and showing its derivative $f'(x) < 0$.
Similarly, considering another function $g(x) = x − \ln(1 + x) − \frac{x^2}{ 2(x + 1)}$ and showing its $g'(x) > 0$. 
Is there some other method that is not lengthy?

Note that
$$ \int_0^x\frac{t}{t+1}dt=x-\ln(x+1). $$
Since $f(t)=\frac{1}{t+1}$ is decreasing in $[0,x]$, one has
$$ \frac{x^2}{2(x+1)}=\int_0^x\frac{t}{x+1}dt<\int_0^x\frac{t}{t+1}dt<\int_0^xtdt=\frac12x^2 $$
and hence
$$ \frac{x^2}{2(x+1)}<x-\ln(x+1) <\frac12x^2, x>0.$$
