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

2 Answers
Mia
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#

Truong-Son N.
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)#

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