help with an integral?

Question

im confused on how to take this integral.
what do i set as u?

S (ln squareroot(x))/x dx
how do i do this

Answer 1

Integral ( ln (sqrt(x))/x dx)

First, it is recommended we use the log properties. But not before changing sqrt(x) into x^(1/2).

Integral ( ln (x^(1/2))/x dx)

As per the log property log[base b](a^c) = c log[base b](a), we can move the (1/2) down outside of the log as a constant.

Integral ( (1/2) ln(x)/x dx)

Pull the constant (1/2) out of the integral, to get

(1/2) Integral ( ln(x)/x dx)

I'm going to rearrange this, to make the next step more obvious.

(1/2) Integral ( ln(x) (1/x) dx )

Here's where we use substitution.

Let u = ln(x).
du = (1/x) dx. {Note: (1/x) dx is the tail end of this integral; that means the tail end of our substituted integral will be du.}

(1/2) Integral ( u du )

Which we can now integrate normally, as

(1/2) (1/2)u^2 + C

(1/4)u^2 + C

But u = ln(x), so our final answer is

(1/4) [ln(x)]^2 + C

Answer 2

Remember that the sqrt(x)/x = 1/sqrt(x) = x^(-1/2)

So I get S[ x^(-1/2) dx]

From here, the integration should be fairly trivial.

Answer 3

ln[sqrt(x)]
=ln[x^(1/2)]
=(1/2)ln[x]

=>(1/2) S [ln x]/x
let S [ln x]/x = A

let u=ln x => du/dx = 1/x
let dv/dx=1/x => v = ln x

=> A = (ln x)^2 - S [ln x]/x
=> A = (ln x)^2 - A
=> 2A = (ln x)^2
=> A = [(ln x)^2]/2

=> S [ln sqrt(x)]/x dx = [(ln x)^2]/4