What is the indefinite integral of #{ln(x)}^2#?

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Answer 1 · 2016-07-03T20:08:07

#= x ln^2 x -2 x ln x +2 x + C#

#int dx qquad ln^2(x)#

we use IBP - #int u v' = uv - int u'v#

#u = ln^2(x), u' = 2 ln x *1/x#
#v' = 1, v = x#

#implies x ln^2 x - int dx qquad 2 ln x#

#implies x ln^2 x -2 color{red}{ int dx qquad ln x} qquad circ#

for the red bit again we IBP

here

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

#implies int dx qquad ln x = x ln x - int dx qquad x*1/x#
# = x ln x - int dx #

# = x ln x - x + C qquad square#

putting #square# into #circ#

#implies x ln^2 x -2 (x ln x - x ) + C#

#= x ln^2 x -2 x ln x +2 x + C#