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}