1) I called initial integral as #I#.
2) I used #x=tany# and #dx=(secy)^2*dy# transforms.
3) I used #tany+coty=2/(sin2y)# identity.
4) I called second integral as #J#.
5) I used #z=2y# and #dz=2dy# transformations in #J#.
6) I used symmetry of the sine function about #pi/2#:
#int_0^(pi) f(sinz)*dz#=#int_0^((pi)/2) 2f(sinz)*dz#
7) I used #int_0^((pi)/2) f(z)*dz#=#int_0^((pi)/2) f((pi)/2-z)*dz# identity.
8) I collected 2 #J#'s.
9) I used #logm+logn=log(m*n)# and #log(a/b)=loga-logb# identities.
10) I found value of #J#.
11) I found value of #I#.
#I#=#int_0^oo log(x+1/x)*dx/(1+x^2)#
After using #x=tany# and #dx=(secy)^2*dy# transforms,
#I#=#int_0^(pi/2) log(tany+coty)*dy#
=#int_0^(pi/2) log(2/(sin2y))*dy#
=#(pi)/2*log2#-#int_0^(pi/2) log(sin2y)*dy#
I called new integral as #J#,
#J#=#int_0^(pi/2) log(sin2y)*dy#
After using #z=2y# and #dz=2dy# transformations,
#J#=#int_0^pi 1/2*log(sinz)*dz#
After using symmetry of the sine function about #pi/2#,
#J#=#int_0^(pi/2) 1/2*2*log(sinz)*dz#
=#int_0^(pi/2) log(sinz)*dz#
=#int_0^(pi/2) log[sin(pi/2-z)]*dz#
=#int_0^(pi/2) log(cosz)*dz#
After collecting 2 integrals,
#2J#=#int_0^(pi/2) log(sinz)*dz#+#int_0^(pi/2) log(cosz)*dz#
=#int_0^(pi/2) log(sinz*cosz)*dz#
=#int_0^(pi/2) log((sin2z)/2)*dz#
=#int_0^(pi/2) log(sin2z)*dz#-#(pi)/2*log2#
=#J#-#(pi)/2*log2#
Hence #J=-(pi)/2*log2#
Thus,
#I#=#int_0^oo log(x+1/x)*dx/(1+x^2)#
=#(pi)/2*log2-J#
=#(pi)/2*log2-(-(pi)/2*log2)#
=#pi*log2#
