HHow do I integrate #f(x)= x^2 e^(-5x)# by parts?

1 Answer
Gió
Jan 28, 2015

Here you have to integrate by parts twice to "kill" your #x^2#.
You start with the first "by parts" as:
#intx^2e^(-5x)dx=x^2*e^(-5x)/(-5)-inte^(-5x)/(-5)*2xdx=#
Which can be written as:
#=x^2*e^(-5x)/(-5)+1/5inte^(-5x)*2xdx=#
Now you go for another integration by parts and get:
#=x^2*e^(-5x)/(-5)+1/5[e^(-5x)/(-5)*2x-inte^(-5x)/(-5)*2dx]=#
#=x^2*e^(-5x)/(-5)+1/5[e^(-5x)/(-5)*2x+2/5inte^(-5x)dx]=#
#=x^2*e^(-5x)/(-5)+1/5e^(-5x)/(-5)*2x+1/5*2/5e^(-5x)/(-5)=#

#=e^(-5x)[-x^2/5-(2x)/25-2/125]+c#

Related questions
See all questions in Integration by Parts
Impact of this question
1634 views around the world
You can reuse this answer
Creative Commons License