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

Question

Christopher Allers · Accepted Answer

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

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#

William Abdalla · Answer

undefined To integrate $ \int \frac{y}{e^{2y}} $ using integration by parts, let $ u = y $ and $ dv = \frac{1}{e^{2y}} dy $. Then, differentiate $ u $ to get $ du = dy $ and integrate $ dv $ to get $ v = -\frac{1}{2e^{2y}} $.

Now, apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the values of $ u $, $ dv $, $ v $, and $ du $ into the formula:

$$ \int \frac{y}{e^{2y}} \, dy = -\frac{y}{2e^{2y}} - \int -\frac{1}{2e^{2y}} \, dy $$

$$ = -\frac{y}{2e^{2y}} + \frac{1}{4} \int e^{-2y} \, dy $$

Now, integrate $ \int e^{-2y} $ to get:

$$ = -\frac{y}{2e^{2y}} + \frac{1}{4} \left( -\frac{1}{2} e^{-2y} \right) + C $$

Where $ C $ is the constant of integration.

Therefore, the integral $ \int \frac{y}{e^{2y}} \, dy $ is:

$$ = -\frac{y}{2e^{2y}} - \frac{1}{8e^{2y}} + C $$