Integral steps using partial integration

Question

I would like to ask steps for below integration.

$v = \int (x-\alpha\beta) x^{\alpha - 1} e^{-\frac{x}{\beta}} dx$

I have the final answer:

$v = -\beta x^\alpha e^{-\frac{x}{\beta}} +c$

But I confused how is the steps. I only know we need to do partial integration.

I try to simplify the form as

$\int x^\alpha e^{-\frac{x}{\beta}} dx - \int \alpha\beta x^{\alpha -1}e^{-\frac{x}{\beta}}dx$

But this seems complicated as well.

Thank you

Answer 1

$\int(x-\alpha\beta)x^{\alpha-1}e^{-\frac{x}{\beta}}~dx$

$=\int x^\alpha e^{-\frac{x}{\beta}}~dx-\int\alpha\beta x^{\alpha-1}e^{-\frac{x}{\beta}}~dx$

$=\int x^\alpha e^{-\frac{x}{\beta}}~dx-\int\beta e^{-\frac{x}{\beta}}~d(x^\alpha)$

$=\int x^\alpha e^{-\frac{x}{\beta}}~dx-\beta x^\alpha e^{-\frac{x}{\beta}}+\beta\int x^\alpha~d(e^{-\frac{x}{\beta}})$

$=\int x^\alpha e^{-\frac{x}{\beta}}~dx-\beta x^\alpha e^{-\frac{x}{\beta}}-\int x^\alpha e^{-\frac{x}{\beta}}~dx$

$=-\beta x^\alpha e^{-\frac{x}{\beta}}+C$

Answer 2

One option is to integrate the first term by parts:

$$\int x^\alpha e^{-\frac{x}{\beta}} dx=-\beta x^\alpha e^{-\frac{x}{\beta}}+\alpha\beta\int x^{\alpha-1}e^{-\frac{x}{\beta}}dx.$$

Then introduce the second term...