∫(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