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

Question

1069 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-27T08:42:44

#= 2sqrt{x} (ln (x) - 2 ) + C#

#int \ \ d(ln(x)) / (x^(1/2))x#

we can try IBP

Wikipedia

#u = ln (x), u' = 1/x#
#v' = x^{-1/2}, v = 2x^{1/2}#

so we have

#2x^{1/2} ln (x) - int \ 2x^{1/2}*1/x \ dx#

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

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

#= 2sqrt{x} (ln (x) - 2 ) + C#