The given function is
#(ln x) ^ 2 / x ^ 2#
We are to find out #I = int (ln x) ^ 2 / x ^ 2dx#
Let
#ln x =u =>x=e^u#
Differentiating w.r.t x we have
#(d(x))/(dx) =(d(e^u))/(du)*(du)/(dx) #
#=>1 =e^u*(du)/(dx) #
#=>dx=e^udu#
Now changing the variable the integral becomes
#I=intu^2/(e^(2u))*e^udu=intu^2*e^-udu#
using integration by parts
#I= u^2inte^-udu-int((d(u^2))/(du)inte^-udu)du#
#= u^2inte^-udu-int2u(-e^-u)du#
#= u^2inte^-udu+int2ue^-udu#
#= u^2inte^-udu+2intue^-udu#
#= u^2inte^-udu+2uinte^-udu-2int((du)/(du)inte^-udu)du#
#= u^2inte^-udu+2uinte^-udu-2int(-e^-u)du#
#= u^2inte^-udu+2uinte^-udu+2int(e^-u)du#
#=-u^2e^-u-2ue^-u-2e^-u+c#
Inserting # u = lnx and e^-u =1/x#
#I=-(ln x)^2/x-(2ln x)/x-2/x+c#
