Integral of $$$x \left(x + 1\right)^{\alpha}$$$ with respect to $$$x$$$

Find $$$\int x \left(x + 1\right)^{\alpha}\, dx$$$.

This integral does not have a closed form:

$${\color{red}{\int{x \left(x + 1\right)^{\alpha} d x}}} = {\color{red}{\left(\frac{x^{2} {{}_{2}F_{1}\left(\begin{matrix} 2, - \alpha \\ 3 \end{matrix}\middle| {- x} \right)}}{2}\right)}}$$

Therefore,

$$\int{x \left(x + 1\right)^{\alpha} d x} = \frac{x^{2} {{}_{2}F_{1}\left(\begin{matrix} 2, - \alpha \\ 3 \end{matrix}\middle| {- x} \right)}}{2}$$

Add the constant of integration:

$$\int{x \left(x + 1\right)^{\alpha} d x} = \frac{x^{2} {{}_{2}F_{1}\left(\begin{matrix} 2, - \alpha \\ 3 \end{matrix}\middle| {- x} \right)}}{2}+C$$

$$$\int x \left(x + 1\right)^{\alpha}\, dx = \frac{x^{2} {{}_{2}F_{1}\left(\begin{matrix} 2, - \alpha \\ 3 \end{matrix}\middle| {- x} \right)}}{2} + C$$$A