How do you integrate # ln x / x^(1/2)#?

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Answer 1 · 2016-07-05T17:00:06

#= 2 sqrt x ln x - 4 sqrt x + C#

Use IBP

#int u v' = uv - int u'v#

here

#u = ln x, u' = 1/x#
#v' = x^{-1/2}, v = 2 x^{1/2}#, using the power rule

so we have

#2 sqrt x ln x - 2 int dx qquad 1/x sqrt x#

#= 2 sqrt x ln x - 2 int dx qquad x^{-1/2}#

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

#= 2 sqrt x ln x - 4 sqrt x + C#