Solve for $$\int_0^1 t^{s+a-1}(1-t)^{a-1}(2t-1)^{-1}dt$$ I first convert $(2t-1)^{-1}$ to its binomial and then geometric form following solve for (2t-1)^-1 in binomial and hypergeometric form
$$-\sum_{n=0}^{\infty}\frac{(1)_n2^n}{n!}\int_0^1t^{s+a+n-1}(1-t)^{a-1}dt$$
Notice the integral is a beta function $B(s+a+n,a)$, so the integral then becomes
$$\frac{\Gamma(s+2a)}{\Gamma(s+a)\Gamma(a)}\int_0^1 t^{s+a-1}(1-t)^{a-1}(2t-1)^{-1}dt$$
We know the hypergeometric integral is $$_2F_1(a;b;c;x)=\frac{\Gamma(c)}{\Gamma(c-a)\Gamma(a)}\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}dt$$
So I then have $$-_2F_1(s+a;1;2a+s;2)=\frac{\Gamma(s+2a)}{\Gamma(s+a)\Gamma(a)}\int_0^1 t^{s+a-1}(1-t)^{a-1}(2t-1)^{-1}dt$$