How do you integrate #(ln(x)*(1/x))dx#?

Question

11988 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-03T17:03:11

# = 1/2 (ln x)^2 + C#

#int dx qquad ln(x)*1/x#

you can do this by inspection as #(ln x) ' = 1/x# so we can trial #alpha (ln x)^2 # as a solution.

So #(alpha (ln x)^2 + C)' = 2 alpha ln x 1/x implies 2 alpha = 1, alpha = 1/2#

if you don't fancy that you could use IBP : #int uv' = uv - int u'v#

#u = ln x, u' = 1/x#
#v' = 1/x, v = ln x#

#implies color{red}{int dx qquad ln(x)1/x} = (ln x)^2 - color{red}{int dx qquad ln x 1/x }+ C#

#implies 2 int dx qquad ln(x)1/x = (ln x)^2 + C#

#implies int dx qquad ln(x)1/x = 1/2 (ln x)^2 + C#