I would like to prove the left side to right hand side which is in form of hypergeometeric function. Looking for your hints, suggestions and solultions.
$$ \alpha_{1} \int_{0}^{1} (1-z)^{\alpha_{1}+\alpha_{2}-1}(1+z)^{\alpha_{2}}dz= \frac{1}{\alpha_{1}+\alpha_{2}} {}_2F_1(1-\alpha_{2};2+\alpha_{1}+\alpha_{2}:-1) $$ where $ 2F1(a_{1},a_{2};b_{1};x)= \sum_{i=0}^{\infty} \frac{(a_{1})_{i}(a_{2})_{i}}{(b_{1})_{i}} x_{i}$ is a hypergeometric function
Regards
Using binomial expansion of $(1+z)^{\alpha_2}$ and with changing the order of summation and integral, we find
\begin{align} \int_0^1(1-z)^{a+b-1}(1+z)^b\ dz &= \int_0^1(1-z)^{a+b-1}\sum_{k=0}^{\infty}\dfrac{\Gamma(-b+k)}{\Gamma(-b)}\dfrac{(-z)^k}{\Gamma(1+k)}\ dz\\ &= \sum_{k=0}^{\infty}\dfrac{\Gamma(-b+k)}{\Gamma(-b)}\dfrac{(-1)^k}{\Gamma(1+k)}\int_0^1(1-z)^{a+b-1}z^k\ dz\\ &= \sum_{k=0}^{\infty}\dfrac{\Gamma(-b+k)}{\Gamma(-b)}\dfrac{(-1)^k}{\Gamma(1+k)}{\bf B}(a+b,k+1)\\ &= \sum_{k=0}^{\infty}\dfrac{\Gamma(-b+k)}{\Gamma(-b)}\dfrac{\Gamma(a+b)\Gamma(1+k)}{\Gamma(1+k)\Gamma(a+b+1+k)}\ (-1)^k\\ &= \dfrac{1}{a+b}\sum_{k=0}^{\infty}\dfrac{\Gamma(-b+k)\Gamma(k+1)\Gamma(a+b+1)}{\Gamma(-b)\Gamma(1)\Gamma(a+b+1+k)}\ \dfrac{(-1)^k}{k!}\\ &= \dfrac{1}{a+b} {}_2F_1(-b,1,a+b+1;-1) \end{align} where ${\bf B}(.,.)$ is beta function.