#int" "ye^(0.2y)dy#?

1 Answer
Cesareo R.
Sep 27, 2017

See below.

Explanation:

#d/(dy)(y e^(alpha y)) = alpha y e^(alpha y) + e^(alpha y)#

then

#int y e^(alpha y) dy = 1/alpha y e^(alpha y) - 1/alpha int e^(alpha y) dy #

or

#inty e^(alpha y) dy = 1/alpha(y - 1/alpha)e^(alpha y) + C#

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