How do you integrate #int y/(e^(2y))# by integration by parts method?

1 Answer
Dec 9, 2016

#I = c -(ye^(-2y))/2 - e^(-2y)/4#

Explanation:

We have

#I = int y/e^(2y)dy#

With some simple rewriting, based on the properties of exponentials #1/a^(x) = a^(-x)#

#I = int ye^(-2y)dy#

If we say

#u = y# so #du = dy#
#dv = e^(-2y)dy# so #v = -e^(-2y)/2#

By the integration by parts formula

#int udv = uv - int vdu#

#I = -(ye^(-2y))/2 - int-e^(-2y)/2dy#

Putting the constants outside of the integral

#I = -(ye^(-2y))/2 +1/2int e^(-2y)dy#
#I = c -(ye^(-2y))/2 - e^(-2y)/4#