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

Question

4121 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

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

$= int u^5 \ du$

$= u^6/6 + C$

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

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