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}

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.
f(x) is increasing from 0 1o 1, where it reaches its global maximum, and decreasing from 1 to $\infty$. It is concave from 0 to 2.00841388..., and convex beyond that (note that the 2nd derivative of f(x) reaches its maximum at 3.01890145886..., after which it declines to 0 in the limit as x goes to $\infty$). Hence quasi-concave.
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: 