Asked by marsha
evaluate the integral:
y lny dy
i know it's integration by parts but i get confused once you have to do it the second time
Leibnitz rule (a.k.a. product rule):
d(fg) = f dg + g df
y lny dy = d[y^2/2 ln(y)] - y/2 dy
---->
Integral of y lny dy =
y^2/2 ln(y) - y^2/4 + const.
Instead of partial integration you can use this trick:
Integral of y^a dy = y^(a+1)/(a+1) (1)
Differentiate both sides w.r.t. a:
Integral of y^a Ln(y) dy =
y^(a+1)Ln(y)/(a+1) -y^(a+1)/(a+1)^2 (2)
uUp to an integration constant)
Substitute a = 1 to obtain the answer. Note that the integration constant we could have added to (1) can still depend on the parameter a. So, if you differentiate it you get an integration constant in (2)
y lny dy
i know it's integration by parts but i get confused once you have to do it the second time
Leibnitz rule (a.k.a. product rule):
d(fg) = f dg + g df
y lny dy = d[y^2/2 ln(y)] - y/2 dy
---->
Integral of y lny dy =
y^2/2 ln(y) - y^2/4 + const.
Instead of partial integration you can use this trick:
Integral of y^a dy = y^(a+1)/(a+1) (1)
Differentiate both sides w.r.t. a:
Integral of y^a Ln(y) dy =
y^(a+1)Ln(y)/(a+1) -y^(a+1)/(a+1)^2 (2)
uUp to an integration constant)
Substitute a = 1 to obtain the answer. Note that the integration constant we could have added to (1) can still depend on the parameter a. So, if you differentiate it you get an integration constant in (2)
Answers
Answered by
Bot
as well.
Integral of y lny dy =
y^2/2 ln(y) - y^2/4 + const.
Integral of y lny dy =
y^2/2 ln(y) - y^2/4 + const.
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