Question

What is the integral of `(lnsqrtx)/x`?

Answer 1

`(lnsqrtx)^2+C`

Explanation:

Let`" " color (blue)(u (x)=lnsqrtx`
`" "`
`(du (x))/dx=((sqrtx)')/sqrtx`
`" "`
`rArr (du (x))/dx=(1/(2sqrtx))/(sqrtx)`
`" "`
`rArr (du (x))/dx=(1/(2x))`
`" "`
`rArr(2du (x))/dx=1/x`
`" "`
`rArrcolor (red)(2 (du (x))=1/xdx`
`" "`
`int (ln (sqrtx)/x)dx`
`" "`
`=int lnsqrtx xx 1/xdx`
`" "`
`=intcolor (blue)(u (x))xx color (red)(2 (du (x))`
`" "`
`=2intu (x)du (x)`
`" "`
`=2 (u (x))^2/2+C`
`" "`
`=(u (x))^2+C" "`where C is a constant
`" "`
`=color (blue)((lnsqrtx))^2+C`

Answer 2

`int (lnsqrtx)/xdx = 1/2 int(lnx)/xdx = 1/4ln^2x + C`

or, `ln^2(sqrtx) + C`.

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You were on the right track. It is more simple than it seems; we know that `lna^b = blna`, so `lnsqrtx = lnx^"1/2" = 1/2lnx`. Therefore:

`int lnsqrtx/xdx`

`= 1/2int (lnx)/xdx`

Let `u = lnx` so that `du = 1/xdx`. Therefore:

`=> 1/2 int udu`

`= 1/2 u^2/2 + C`

`= 1/2 ln^2x/2 + C`

`= color(blue)(1/4 ln^2x + C)`

Or, we could rewrite this as:

`= (1/2lnx)^2 + C`

`= color(blue)(ln^2(sqrtx) + C)`