First let's solve the indefinite integral #intln^2(x)/sqrt(x)dx#. by applying integration by parts twice:
First Integration by Parts
Let #u = ln^2(x)# and #dv = 1/sqrt(x)dx#
Then #du = (2ln(x))/x# and #v = 2sqrt(x)#
Thus
#intln^2(x)/sqrt(x)dx = intudv#
#= uv - intvdu#
#=2sqrt(x)ln^2(x) - 4intln(x)/sqrt(x)dx#
Second Integration by Parts
Let #u = ln(x)# and #dv = 1/sqrt(x)#
Then #du = 1/x# and #v = 2sqrt(x)#
Thus
#intln(x)/sqrt(x)dx = intudv#
#= uv - intvdu#
#= 2sqrt(x)ln(x) - 2int1/sqrt(x)dx#
#= 2sqrt(x)ln(x) - 4sqrt(x) + C#
Putting it together, we get
#intln^2(x)/sqrt(x)dx = 2sqrt(x)ln^2(x) - 4intln(x)/sqrt(x)dx#
#= 2sqrt(x)ln^2(x) - 4(2sqrt(x)ln(x) - 4sqrt(x) + C)#
#= 2sqrt(x)(ln^2(x) - 4ln(x) + 8) + C#
Now we can evaluate the original definite integral.
#int_0^piln^2(x)/sqrt(x)dx = [2sqrt(x)(ln^2(x) - 4ln(x) + 8)]_0^pi#
(Note that as there is a discontinuity at #0# we must evaluate this using a limit)
#= 2sqrt(pi)(ln^2(pi) - 4ln(pi) + 8) - lim_(x->0)2sqrt(x)(ln^2(x) - 4ln(x) + 8)#
#= 2sqrt(pi)(ln^2(pi) - 4ln(pi) + 8) - 0#
#= 2sqrt(pi)(ln^2(pi) - 4ln(pi) + 8)#
