How do you integrate #ln(x)/x^2#?

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Answer 1 · 2015-06-29T05:51:48

This can be rewritten as:

#int lnx * 1/x^2dx#

Using integration by parts, let:
#u = lnx#
#du = 1/xdx#
#dv = 1/x^2dx#
#v = -1/xdx#

With the formula
#uv - intvdu#

#=> -lnx/x - int -1/x*1/xdx#

#= -lnx/x - int -1/x^2dx#

#= color(blue)(-lnx/x - 1/x + C)#