How do you integrate #int (lnx)^5/x# using substitution?

Question

3593 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-26T11:26:07

#= (ln x)^6/6 + C #

for starters, there's a clear pattern here:

#d/dx ( ln^n x) = n ln^(n-1) x 1/x#

so

#int (lnx)^5/x \ dx = int d/dx ( 1/6ln^6 x) \ dx = 1/6ln^6 x + C#

but if you are required to introduce substitution into this, which seems OTT, I suppose you could say that

#u = ln x, du = 1/x dx #

so you have

#int u^5/x \ x \ du #

#= int u^5 \ du #

#= u^6/6 + C #

#= (ln x)^6/6 + C #