Question

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

Answer 1

`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`