What is the antiderivative of #(ln x)^2/x^2#?

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Answer 1 · 2016-08-29T12:13:40

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

#int ln^2 x/x^2 dx#

#= int ln^2 x 1/x^2 dx#

#= int ln^2 x d/dx(- 1/x) dx#

#= -1/x ln^2 x + int d/dx( (ln x)^2 ) 1/x dx#

#= -1/x ln^2 x + int ( 2 ln x 1/x ) 1/x dx#

#= -1/x ln^2 x + 2 int ln x 1/x^2 dx#

#= -1/x ln^2 x + 2 int ln x d/dx( -1/x) dx#

#= -1/x ln^2 x + 2( -1/x ln x + int d/dx( ln x ) 1/x dx) #

#= -1/x ln^2 x + 2( -1/x ln x + int 1/x^2 dx) #

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

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