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

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Answer 1 · 2016-12-09T16:15:32

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

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 $v = -e^(-2y)/2$

$int udv = uv - int vdu$

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

Putting the constants outside of the integral

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

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