<s>[INST] Quasi-concavity of $f(x)=\frac{1}{x+1} \int_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt$ I want to prove that function 
\begin{equation}
f(x)=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt
\end{equation}
is quasi-concave. One approach is to obtain the closed form of the integral (provided below) and then prove that the result is quasi-concave. I tried this but it seems to be difficult. Do you have any idea how to prove the quasi-concavity? 
\begin{align}
f(x)&=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt\\
&= \frac{1}{x+1} \left[ \log(2x + 2) + (2x+1)\log \left(1+\frac{1}{2x+1} \right) -\log(x+2) - (x+1)\log \left( 1+\frac{1}{x+1} \right) \right]
\end{align}
 [/INST] For $x>-1/2$, we have 
$$
f(x)=\ln 4-\frac{x+2}{x+1}\, \ln \frac{x+2}{x+1}
-\frac{2 x+1}{x+1} \ln\frac{2 x+1}{x+1}
$$
and hence 
$$
f'(x)=\frac1{(x+1)^2}\ln \left(1+\frac{1-x}{2 x+1}\right),
$$
which is $>0$ for $x\in(-1/2,1)$ and $<0$ for $x>1$. So, $f$ increases on $(-1/2,1]$ and decreases on $[1,\infty)$. So, $f$ is quasi-concave.
</s>