Find $int \ ln(sqrt(x))/x dx$ ?

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Answer 1 · 2017-09-26T14:43:43

$int \ ln(sqrt(x))/x \ dx = ln^2x/x + C$

We seek:

$I = int \ ln(sqrt(x))/x \ dx$

Using the properties of logarithms, we can write this integral as follows:

$I = int \ 1/2 \ ln(x)/x \ dx$

Which then leads to a suggestive substitution:

Let $u =ln x$ => $(du)/dx = 1/x$

So if we now substitute this into the integral, we get:

$I = 1/2 \ int \ u \ du$

Which is now a standard integral, so we have:

$I = 1/2 u^2/2 + C$

And reversing the substitution:

$I = 1/4 (lnx)^2 + C$

Find int \ ln(sqrt(x))/x dx int \ ln(sqrt(x))/x dx ?