Solve this integral of degree $2043$ 
Find the value of $$\frac{\displaystyle\int_0^1(x+1)^{1010}\:\:dx}{\displaystyle\int_0^1(x^{2043}+1)^{1010}\:\:dx}$$

I evaluated the value of the numerator as $$\frac{2^{1011}-1}{1011}$$ But can't do the same for the denominator.
Any help is greatly appreciated.

Using the Gaussian hypergeometric function
$$\int  \left(x^p+1\right)^n \,dx=x \, _2F_1\left(-n,\frac{1}{p};1+\frac{1}{p};-x^p\right)$$
$$\int_0^1  \left(x^p+1\right)^n \,dx=\, _2F_1\left(-n,\frac{1}{p};1+\frac{1}{p};-1\right)$$
$$I_{n,p}=\frac{\displaystyle\int_0^1(x+1)^{n}\:\:dx}{\displaystyle\int_0^1(x^{p}+1)^{n}\:\:dx}=\frac{2^{n+1}-1}{(n+1)\, _2F_1\left(-n,\frac{1}{p};1+\frac{1}{p};-1\right)}$$
What is interesting to notice is that
$$I_{n,2n+k}\sim \frac{2 \,i^{\frac{1}{n}}\, \left(2^{n+1}-1\right) n}{(n+1) \,B_{-1}\left(\frac{1}{2 n},n+1\right)}+k+O(k^2)$$ which, for $n=1010$ and $k=23$ gives $2039.00$ while the result is $2038.96$.
