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 $K \in\mathbb{N}$.

Let $u=1-m$. Then:
$$I=\frac{\Gamma(K+1)}{\Gamma(K)}\int_{0}^{+\infty}\frac{1}{(1+x)(1+x-u)^K}\,dx=K\int_{0}^{+\infty}\frac{\left(1-\frac{u}{x+1}\right)^{-K}}{(1+x)^{K+1}}\,dx$$
but since:
$$(1-z)^{-K} = \sum_{n=0}^{+\infty}\binom{n+K-1}{K-1} x^n$$
it follows that:
$$ I = K\sum_{n=0}^{+\infty}u^n\binom{n+K-1}{K-1}\int_{0}^{+\infty}\frac{dx}{(1+x)^{K+n+1}}=\sum_{n=0}^{+\infty}u^n\binom{n+K-1}{K-1}\frac{K}{K+n}.$$
If we set $A_n = \binom{n+K-1}{K-1}\frac{K}{K+n}$ then we have:
$$\frac{A_{n+1}}{A_n}=\frac{(n+K)^2}{(n+1)(n+K+1)}$$
from which it follows that:
$$ I = \phantom{}_2 F_1(K,K;K+1;u) = \phantom{}_2 F_1(K,K;K+1;1-m)$$
as wanted.
