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

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Answer 1 · 2016-07-16T13:39:25

#I = 1/2 ln^2 x + C#

#I = int \ ln x / x \ dx#

this can be done by inspection as its #int \ f(x) f'(x) \ dx =1/2f^2(x) +C#

or we can use another method eg IBP......#int u v' = uv - int u' v#

#I = int \ ln x / x \ dx#

#= int d/dx(ln x) \ ln x \ dx#

#= ln x ln x - int \ \ ln x \ d/dx(ln x) \ dx#

#= ln^2 x - int \ \ ln x/x \ dx#

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

#I = 1/2 ln^2 x + C#