Question

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

Answer 1

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

Explanation:

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`