∫(lnx)^ndx=x(lnx)^(n)-n∫(lnx)^(n-1)dx;

Question

All I need to know is what dv=

I already found u=(ln(x))^n

du= n(ln(x))^(n-1)*(1/x)

Answer 1

Using Integration by Parts with u = (lnx)ⁿ and dv = dx

So, du = n(lnx)^n-1(1/x)dx and v = ∫dv = x

Then, ∫(lnx)ⁿdx = uv - ∫vdu = x(lnx)ⁿ - ∫xn(lnx)^n-1(1/x)dx

= x(lnx)ⁿ - n∫(lnx)^n-1dx

Answer 2

In the problem you've worked out, dv = dx or, you could say dv = 1dx