What is the integral of (lnsqrtx)/x?

2 Answers
May 24, 2017

(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

May 24, 2017

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

or, ln^2(sqrtx) + C.


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)