Wolfram Alpha provides following answer for the integral $\displaystyle\int \frac{xe^{-x}}{(ax-c)^N}dx\,$:
$$-e^{-\frac{c}{a}}a^{-N-1}\left(c\Gamma(1-N,x-\frac{c}{a})+a\Gamma(2-N,x-\frac{c}{a})\right)+Constant$$
I do not understand how does incomplete Gamma function comes in the answer. As far as I know the definition of incomplete Gamma function is some definite integral but how does it appears in the indefinite integral. I will be very thankful if somebody guide me how to reach to the above answer and how does the limits appear to give Gamma function in the answer.
Thanks in advance.
Substituting $u = ax-c$ gives us $$I = \frac {e^{-c/a}}{a^2} \int \frac {(u+c)e^{-u/a}}{u^N} \mathrm {d}u = \frac {e^{-c/a}}{a^2} I_1$$
Now $$I_1 = c\int \frac {e^{-u/a}}{u^N} \mathrm {d}u + \int u^{1-N} e^{-u/a} \mathrm {d}u = c I_{11} + I_{12} $$
In $I_{11}$, substituting $v = (u/a)^{1-N} $, gives us $$I _{11} = \frac {1}{1-N} \int e^{-v^{\frac {1}{1-N}}} \mathrm {d}v$$ Now this is an incomplete gamma function whose value comes out to be $$I_{11}= \frac {1}{1-N} [-(1-N)\Gamma (1-N, v^{\frac {1}{1-N}})] $$
Also, $I_{12} $ can be solved similarly writing $w= (u/a)^{2-N}$ giving us the final result as $$I_{12} = -a^{2-N} \Gamma (2-N, v^{\frac {1}{2-N}}) $$
ADD both of them and manipulate to get the final result. Hope it helps.