#A=int_1^oo (1/u^2-1/u^4)*(du)/Lnu#
After using #x=Lnu#, #u=e^x# and #du=e^x*dx# transforms, #A# became,
#A=int_0^oo [1/(e^x)^2-1/(e^x)^4]*(e^x*dx)/x#
=#int_0^oo (e^(-2x)-e^(-4x))*(e^x*dx)/x#
=#int_0^oo ((e^(-x)-e^(-3x))*dx)/x#
Now, I solved this integral via Feynman's trick by choosing #b=3#
#A(b)=int_0^oo ((e^(-x)-e^(-bx))*dx)/x#
After differentiation both sides according to #b#
#A'(b)=int_0^oo e^(-bx)*dx#
=#[-1/b*e^(-bx)]_0^oo#
=#1/b#
After integrating both sides,
#A(b)=Lnb+C#
Now I chose a strategic value #b = b_0# in order to make our integrand vanish so that #A(b_0)=0#. In this case, take #b=1# so that #A(1)=0#. Consequently,
#Ln1+C=0# or, #C=0#
Hence, I found value of #A# after setting #b=3#
#A=A(3)=ln3#
Thus value of #e^A# is #e^(Ln3)=3#
