How do you integrate #int ln^(6)x/x dx#?

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Answer 1 · 2015-06-08T00:41:53

First, notice that you can rewrite this as:

#int ln^6x/xdx = int (lnx)^6*1/xdx#

Now you can see that we can do u-substitution.

Let:
#u = lnx#
#du = 1/xdx#

#int (lnx)^6*1/xdx = intu^6du#

#= u^7/7 + C#

#= ln^7x/7 + C#