Hypergeometric function integral representation How to prove the following relation?
$$ \, _2{F}_1(K,K;K+1;1-m) = \frac{\Gamma (K+1)}{\Gamma (K)} \int_0^{\infty } \frac{1}{(1+x) (m+x)^K} \, dx $$
where $_2{F}_1(.,.;.;.)$ is the hypergeometric function, $m\in\mathbb{R}^+$ , and $K \in\mathbb{N}$.

This is to answer the question which @sky-light asked in the comment section.
Based on the formula from the original question:
$\frac{_2F_1(K+1,K+1;K+2;1-m)}{K+1} = \int_{0}^{\infty} \frac{dx}{(1+x)(m+x)^{K+1}} = \frac{1}{1-m} \int_{0}^{\infty} \frac{(1+x)-(m+x)}{(1+x)(m+x)^{K+1}}dx = \frac{1}{1-m} \big[ \int_{0}^{\infty}  \frac{dx}{(m+x)^{K+1}} - \int_{0}^{\infty} \frac{dx}{(1+x)(m+x)^k} \big] = \frac{1}{1-m} \big[ -\frac{1}{K(m+x)^K}\big|_0^{\infty} - \frac{_2F_1(K,K;K+1;1-m)}{K}\big] = \frac{m^{-K} - _2F_1(K,K;K+1;1-m)}{K(1-m)}$
Then, the two are equal.
P/S: at the time I posted this answer, my privilege was not enough to answer in the comment.
