∫(lnx)^2*dx

=∫udv

=uv - ∫vdu

=x*(lnx)^2 - ∫x *2lnx/x *dx

=x*(lnx)^2 - 2∫lnx*dx

Reuse the partial integral for ∫lnx*dx, let m = lnx and dn = dx, then dm = dx/x and n = x.

∫lnx*dx

=∫mdn

=mn - ∫ndm

=x*lnx - ∫x*dx/x

=x*lnx - x

So, the original integral:

=x*(lnx)^2 - 2x*lnx + 2x + C