1) I called initial integral as I.
2) I used x=tany and dx=(secy)2⋅dy transforms.
3) I used tany+coty=2sin2y 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 π2:
∫π0f(sinz)⋅dz=∫π202f(sinz)⋅dz
7) I used ∫π20f(z)⋅dz=∫π20f(π2−z)⋅dz identity.
8) I collected 2 J's.
9) I used logm+logn=log(m⋅n) and log(ab)=loga−logb identities.
10) I found value of J.
11) I found value of I.
I=∫∞0log(x+1x)⋅dx1+x2
After using x=tany and dx=(secy)2⋅dy transforms,
I=∫π20log(tany+coty)⋅dy
=∫π20log(2sin2y)⋅dy
=π2⋅log2-∫π20log(sin2y)⋅dy
I called new integral as J,
J=∫π20log(sin2y)⋅dy
After using z=2y and dz=2dy transformations,
J=∫π012⋅log(sinz)⋅dz
After using symmetry of the sine function about π2,
J=∫π2012⋅2⋅log(sinz)⋅dz
=∫π20log(sinz)⋅dz
=∫π20log[sin(π2−z)]⋅dz
=∫π20log(cosz)⋅dz
After collecting 2 integrals,
2J=∫π20log(sinz)⋅dz+∫π20log(cosz)⋅dz
=∫π20log(sinz⋅cosz)⋅dz
=∫π20log(sin2z2)⋅dz
=∫π20log(sin2z)⋅dz-π2⋅log2
=J-π2⋅log2
Hence J=−π2⋅log2
Thus,
I=∫∞0log(x+1x)⋅dx1+x2
=π2⋅log2−J
=π2⋅log2−(−π2⋅log2)
=π⋅log2
