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}

The derivative is:  ${d f(x) \over d x} = \frac{-\log (x+1)+\log (x+2)+\log \left(\frac{x+1}{2 x+1}\right)}{(x+1)^2}$, which is:

and hence is both positive and negative, as Mark L. Stone wrote.
