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

1 Answer
Aug 21, 2016

#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#