Question

How do you find the integral of `f(x)=x^(n-1)e^x` using integration by parts?

Answer 1

`int x^n e^x dx = e^x sum_{k=0}^n (-1)^k((n!)/(k!))x^k`

Explanation:

`d/(dx)(x^n e^x)=n x^{n-1}e^x+x^n e^x`

then

`int x^n e^x dx + n int x^{n-1}e^x dx = x^n e^x`

now calling `I_n=e^{-x}int x^n e^x dx` we have a recurrence equation

`I_n +nI_{n-1} = x^n` with condition

`I_0 = e^{-x}int e^xdx = 1`

Solving this recurrence equation is straightforward as can be seen in
https://en.wikipedia.org/wiki/Recurrence_relation in

Solving first-order non-homogeneous recurrence relations with variable coefficients

giving as result

`I_n = sum_{k=0}^n (-1)^k((n!)/(k!))x^k`

but `I_n=e^{-x}int x^n e^x dx`

so

`int x^n e^x dx = e^x sum_{k=0}^n (-1)^k((n!)/(k!))x^k`