Question

How do you evaluate the integral of `int 3x^5 e^x dx`?

Answer 1

`3(x^5 e^x - 5( x^4 e^x - 4( x^3 e^x-3( x^2e^x -2 (x-1)e^x))))+C`

or

`3 ( x^5 - 5 x^4 + 20 x^3 - 60 x^2 + 120 x - 120)e^x + C`

Explanation:

In this development we will consider the integration constant `C = 0`.
Calling `I_n = int x^n e^x dx` and after

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

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

or

`x^n e^x= n I_{n-1}+I_n`

with `I_0 = int x^0 e^x dx = int e^x dx = e^x`
so we can iterate

`I_0+I_1 = x e^x-> I_1 = x e^x - e^x = (x-1)e^x`
`2I_1+I_2 = x^2 e^x->I_2 = x^2e^x -2 (x-1)e^x`
`3I_2+I_3 = x^3 e^x->I_3 = x^3 e^x-3( x^2e^x -2 (x-1)e^x)`
`4I_3+I_4 = x^4 e^x->I_4 = x^4 e^x - 4( x^3 e^x-3( x^2e^x -2 (x-1)e^x))`

and finally

`5I_4+I_5 = x^5 e^x` obtaining

`I_5 = x^5 e^x - 5( x^4 e^x - 4( x^3 e^x-3( x^2e^x -2 (x-1)e^x)))`

Our integral is `3I_5` so

`int 3x^5 e^x dx = 3I_5`