How do you find the integral of #int (lnx / sqrt(x)) dx# from 0 to 1?

2 Answers
Sep 15, 2015

I found: #-4#

Explanation:

Let us try by changing #1/sqrt(x)==x^(-1/2)# and then Integrate By Parts:
#int(ln(x)/sqrt(x))dx==int(x^(-1/2)ln(x))dx=#
#=2x^(1/2)ln(x)-int(2x^(1/2)*1/xdx)=#
#=2x^(1/2)ln(x)-int(2x^(1/2-1)dx)=#
#=2x^(1/2)ln(x)-int(2x^(-1/2)dx)=#
#=2x^(1/2)ln(x)-4x^(1/2)=#
let us use our extrema:
#=2x^(1/2)ln(x)-4x^(1/2)|_0^1##=-4#

Only problematic part is #2x^(1/2)lnx# when #x->0_+#, so we will find the limit:

#lim_(x->0_+)2x^(1/2)lnx= 0*(-oo)# which is undefined.

Using the rule of L'Hospital:

#lim_(x->0_+)2x^(1/2)lnx=2lim_(x->0_+)lnx/x^(-1/2)=#

#=2lim_(x->0_+)(1/x)/(-1/2x^(-3/2))=-4lim_(x->0_+)x^(3/2)/x=#

#=-4lim_(x->0_+)sqrtx=-4*0=0#

So, the above results holds.

Sep 16, 2015

i got -4 by changing variabe method

Explanation:

let the # x = t^2#
then #x^(1/2) = t#
#dx# = #2tdt#
limits:
if x= 0 then t= 0
if x= 1 then t=1
applying integration
#int_0^1# #lnt^2/t#( #2tdt#)
=#int_0^1##4lntdt#
=4#int_0^1##lntdt#
applying by parts
4#lnt#*#int_0^1##dt#-#int_0^1##t*(1/t)##dt#
=4[#ln1# 1- 0 #ln0#-{1-0}]
=4[0-1]
=-4