设 u = (lnx)^2，dv = dx。则 du = 2lnx/x *dx，v = x

∫(lnx)^2*dx

=∫udv

=uv - ∫vdu

=x*(lnx)^2 - ∫x *2lnx/x *dx

=x*(lnx)^2 - 2∫lnx*dx

对于∫lnx*dx 再次使用分部积分法，设 m = lnx，dn = dx，则 dm = dx/x，n = x

∫lnx*dx

=∫mdn

=mn - ∫ndm

=x*lnx - ∫x*dx/x

=x*lnx - x

所以，原积分：

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