Solve $\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$ using hypergeometric functions How can we solve this integral?
$$\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$$
I used geometric series
$$=\int_0^1(x+1)^b\sum_{n=0}^\infty\left(\frac{x}{\alpha}\right)^{n+1}dx=\sum_{n=0}^\infty\int_0^1\left(\frac{x}{\alpha}\right)^{n+1}(x+1)^bdx$$
Using
$$\int_0^1t^b(1-t)^c(1-zt)^adt=\Gamma(b+1)\Gamma(c+1){_2}F_1(-a,b+1;c+b+2;z)$$
I got $$\sum_{n=0}^\infty a^{-n-1}\Gamma(n+2){_2}F_1(-b,n+2;n+3;-1)$$
The problem is I don't know how to simplify this any further. How can ${_2}F_1(-b,n+2;n+3;-1)$ be simplified?

Answer for $\alpha> 1$ and $b\notin\Bbb Z$
For $\alpha>1$, we can write the integral as
$$
I=\int_0^1\frac{x(x+1)^b}{\alpha-x}dx=\int_0^1\frac{x(x+1)^b}{\alpha+1-(x+1)}dx
=
\frac{1}{\alpha+1}\int_0^1\frac{x(x+1)^b}{1-\frac{x+1}{\alpha+1}}dx,
$$
where $0<\frac{1}{\alpha+1}\le\frac{x+1}{\alpha+1}\le \frac{2}{\alpha+1}<1$. Hence, the term $\frac{1}{1-\frac{x+1}{\alpha+1}}$ can be expanded by its Taylor's series. We can then write the integral as
$$
I{=\frac{1}{\alpha+1}\int_0^1x(x+1)^b\sum_{n=0}^\infty(\frac{x+1}{\alpha+1})^ndx,
\\=
\frac{1}{\alpha+1}\int_0^1x\sum_{n=0}^\infty\frac{(x+1)^{n+b}}{(\alpha+1)^n}dx,
\\=
\frac{1}{\alpha+1}\int_0^1(x+1-1)\sum_{n=0}^\infty\frac{(x+1)^{n+b}}{(\alpha+1)^n}dx,
\\=
\frac{1}{\alpha+1}\int_0^1\sum_{n=0}^\infty\frac{1}{(\alpha+1)^n}[(x+1)^{n+b+1}-(x+1)^{n+b}]dx
\\=
\frac{1}{\alpha+1}\sum_{n=0}^\infty\frac{1}{(\alpha+1)^n}\left[\frac{2^{n+b+2}-1}{n+b+2}-\frac{2^{n+b+1}-1}{n+b+1}\right]
.
}
$$
The value of the integral is $\infty$ for $\alpha=1$ and is not explicitly defined for $0<\alpha<1$, unless through the limit of Cauchy integral. The case for $\alpha<0$ should also follow similarly.
