Asked by omar
integral lnx^2019
Answers
Answered by
oobleck
ln x^2019 = 2019 lnx
so the derivative is 2019/x
Now, if you meant (lnx)^2019 then you have to use integration by parts
u = (lnx)^2019 ... du = 2019/x (lnx)^2018
dv = dx v = x
∫(lnx)^2019 = x (lnx)^2019 - ∫ 2019 (lnx)^2018
you can see the pattern here.
∫(lnx)^n dx =
n
∑ x * n!/(n-k)! (lnx)^(n-k)
k=0
so the derivative is 2019/x
Now, if you meant (lnx)^2019 then you have to use integration by parts
u = (lnx)^2019 ... du = 2019/x (lnx)^2018
dv = dx v = x
∫(lnx)^2019 = x (lnx)^2019 - ∫ 2019 (lnx)^2018
you can see the pattern here.
∫(lnx)^n dx =
n
∑ x * n!/(n-k)! (lnx)^(n-k)
k=0
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