How may one integrate the function
$\frac{x^a}{(bx-1)^2}$ with respect to x?
I have attempted to evaluate the function with Mathematica and it gives the solution in the form of:
$\frac{x^{a+1}._2F_1(1, 2; 2-a; \frac{1}{1-bx})}{(a-1)(bx-1)^2}$
But I have no idea what substitutions have to be made for one to obtain the above solution.
Thanks in advance for the help
This is another possible evaluation of that integral. For $|bx|<1$ \begin{align*}\int \frac{x^a}{(bx-1)^2} dx&=\int x^a\sum_{k=1}^{\infty}k(bx)^{k-1} dx=\sum_{k=1}^{\infty}kb^{k-1}\int x^{a+k-1}\,dx =\sum_{k=1}^{\infty}\frac{kb^{k-1} x^{a+k}}{a+k}+c\\ &=x^{a+1}\sum_{k=1}^{\infty}\frac{k(bx)^{k-1}}{a+k}+c =x^{a+1}\sum_{k=1}^{\infty}(bx)^{k-1}-\frac{ax^{a}}{b}\sum_{k=1}^{\infty}\frac{(bx)^{k}}{a+k}+c\\ &=\frac{x^{a+1}}{1-bx}-\frac{ax^{a}}{b}\Phi(bx,1,a)+\frac{x^{a}}{b}+c\\ &=\frac{x^{a}}{b(1-bx)}-\frac{ax^{a}}{b}\Phi(bx,1,a)+c \end{align*} where $\Phi(z,s,a)$ is the Lerch Transcendent.