Solving the integral $\int\frac{x^{\alpha}}{1+\psi x}\,\mathrm dx$ I try to solve the following integral:
$$\int\frac{x^{\alpha}}{1+\psi x}\,\mathrm dx$$
What I try is to use the substitution method. I define $1+\psi x=u$ and this yields $\mathrm du=\psi\, \mathrm dx$
This gives me at final:
$$\int\frac{\left(u-1\right)^{\alpha}}{u\psi^{2}}\,\mathrm du$$
This one did not help me to solve the integral. What can I do to solve it? 

Let $y=-bx$
\begin{align}
\int \frac{x^{a}}{1+bx} dx &= \left(\frac{1}{-b}\right)^{a+1} \int y^{a} (1-y)^{-1} dy \\
&= \left(\frac{1}{-b}\right)^{a+1} \mathrm{B}_{y}(a+1,0) \\
&= \left(\frac{1}{-b}\right)^{a+1} \mathrm{B}_{-bx}(a+1,0) \\
&= \left(\frac{1}{-b}\right)^{a+1} \frac{(-bx)^{a+1}}{a+1} {}_{2}\mathrm{F}_{1}(a+1,1;a+2;-bx) \\
&= \frac{x^{a+1}}{a+1} {}_{2}\mathrm{F}_{1}(a+1,1;a+2;-bx)
\end{align}
Note:
$$\mathrm{B}_{z}(x,y) = \int\limits_{0}^{z} t^{x-1} (1-t)^{y-1} dt 
= \frac{z^{x}}{x} {}_{2}\mathrm{F}_{1}(x,1-y;x+1;z)$$
the incomplete beta function and Gauss's hypergeometric function respectively.
