What is the antiderivative of #ln(2x)/x^(1/2)#?

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Answer 1 · 2017-02-09T20:55:56

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

#int ln(2x)/x^(1/2) dx#

We set it up for IBP:
#= int ln(2x) d/dx(2 x^(1/2) ) dx#

Applying the IBP:
#= 2 x^(1/2) ln(2x) - int d/dx( ln(2x) )(2 x^(1/2) ) dx#

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

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

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

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