Vtyjkyuk

I am trying to work out the asymptotic approximation (for large $n$) of the following:
$$sum _i=0^n log (binomni+1)$$
So far I tried to write the expression as,
$$sum^n_i=0log(x)=-sum^n_i=0xint^infty_0e^-xtlog tdt-gamma$$
where I took $x=binomni+1$ and $gamma$ is Euler gamma, but I am not sure if this method is so useful. Any other approach is appreciated.

The proposer has pointed out the dominant term and the addition of 1 within the log argument is a minor correction.

$$S:=sum_k=0^nlog(1+binomnk) =sum_k=0^n log(binomnk(1+1/binomnk)sim$$
$$simsum_k=0^n log(binomnk)+sum_k=0^nfrac1binomnk-frac12sum_k=0^nfrac1binomnk^2+...$$
where a Taylor series of $log(1+x)$ has been used.
It can be shown that
$$sum_k=0^nfrac1binomnk =2(1+frac1n+...) text and sum_k=0^nfrac1binomnk^2=2(1+1/n^2+...)$$
We will always know the leading coefficient is 2 because of the $k=0$ and $k=n$ terms. We have an infinite sum of these terms, which fortunately converges because of their alternating weight of $(-1)^m/m.$ The log-binomial sum can be done in closed form:
$$sum_k=0^n log(binomnk)=logBig(fracn!^n+1G(n+2)^2Big). $$
$G$ is the Barne's G and its asymptotic expansion is well-known.
Collecting,
$$Ssim logBig(fracn!^n+1G(n+2)^2Big) + 2log2+frac2n+O(frac1n^2) $$
Use this expansion for $n=20$ and 4 correct digits are obtained.