Pose #y=lnx# and decompose the resulting rational function of #y# in partial fractions:
#(2y-3)/(y^2-3x+2) = (2y-3)/((y-1)(y-2))#
#(2y-3)/(y^2-3x+2) = A/(y-1)+B/(y-2)#
#(2y-3)/(y^2-3x+2) = (A(y-2)+B(y-1))/((y-1)(y-2))#
#2y-3 = (A+B)y - (2A+B)#
#{(A+B =2),(2A+B=3):}#
#{(A=1),(B=1):}#
So:
#int (2lnx-3)/(ln^2x-3lnx+2) dx = int dx/(lnx-1) + int dx/(lnx-2)#
Consider the first integral:
#int dx/(lnx-1)#
and substitute #t=lnx-1#, #x=e^(t+1) #, #dx = e^(t+1)dt#:
#int dx/(lnx-1) = int e^(t+1)/tdt = e int e^t/t dt#
This last integral cannot be expressed in terms of elementary functions but is a specific function called exponential integral, so:
# int dx/(lnx-1) = e Ei(t) +C#
and undoing the substitution:
# int dx/(lnx-1) = e Ei(lnx-1) +C#
Similarly:
# int dx/(lnx-2) = e^2 Ei(lnx-2) +C#
and in conclusion:
#int (2lnx-3)/(ln^2x-3lnx+2) dx =e Ei(lnx-1) + e^2 Ei(lnx-2) +C#
