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 $K \in\mathbb{N}$.

Refer to the integral definition of $ _2F_1(a,b;c;z)$ in the general case :
$$ \, _2{F}_1(a,b;c;z) = \frac{\Gamma (c)}{\Gamma (b) \Gamma (c-b)} \int_0^{\infty} t^{-b+c-1} \, (t+1)^{a-c} \, (t-z+1)^{-a} \, dt $$
In the particular case : $a=K$ ; $b=K$ ; $c=K+1$ ; $z=1-m$ ; $t=x$ 
and with $\Gamma(c-b)=\Gamma(1)=1$  leads to : 
$$ \, _2{F}_1(K,K;K+1;1-m) = \frac{\Gamma (K+1)}{\Gamma (K)} \int_0^{\infty } \frac{1}{(1+x) (m+x)^K} \, dx $$
Note that $\frac{\Gamma(K+1)}{\Gamma(K)}=K$
