You should learn the IBP formula:
# int u(dv)/dxdx=uv - int v (du)/dxdx #
So essentially we are looking for one function that simplifies when it is differentiated, and one that simplifies when integrated (or at least is integrable).
Hopefully you can spot that #lnx# is not easy to integrate (you need to using IBP again), but is simpler when differentiated.
Let # { (u=lnx, => , (du)/dx=1/x), ((dv)/dx=1/x^3=x^-3, =>, v=x^-2/-2 = -1/(2x^2) ) :}#
So IBP gives;
# int lnx 1/x^3 dx = (lnx)(-1/(2x^2) ) - int ( -1/(2x^2))(1/x)dx#
# :. int lnx /x^3 dx = (-lnx)/(2x^2 ) + 1/2 int ( 1/x^3 )dx#
# :. int lnx /x^3 dx = (-lnx)/(2x^2 ) + 1/2 (-1/(2x^2)) + C#
# :. int lnx /x^3 dx = (-lnx)/(2x^2 ) - 1/(4x^2) + C#
