What is the antiderivative of #(ln^6 x)/x#?

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Answer 1 · 2016-01-31T18:22:34

#(ln^7x)/7+C#

Use substitution.

Let #u=lnx#, so #du=1/xdx#.

Finding the antiderivative is equivalent to finding

#int(ln^6x)/xdx#

This can be rewritten as

#=intln^6x(1/x)dx#

Using the substitutions previously defined

#=intu^6du#

This is equal to

#=1/7u^7+C#

Resubstitute #u#:

#=(ln^7x)/7+C#