#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#