) I used tan(u+v)=(tanu+tanv)/(1-tanu*tanv)tan(u+v)=tanu+tanv1−tanu⋅tanv identity for decomposing arctan((2x-1)/(1+x-x^2))arctan(2x−11+x−x2).
2) I used x=1-ux=1−u transform in II integral.
3) After summing 2 integrals, I found result.
II=int_0^1 arctan[(2x-1)/(1+x-x^2)]*dx∫10arctan[2x−11+x−x2]⋅dx
=int_0^1 arctan((2x-1)/(1-x*(x+1)))*dx∫10arctan(2x−11−x⋅(x+1))⋅dx
=int_0^1 arctanx*dx∫10arctanx⋅dx+int_0^1 arctan(x-1)*dx∫10arctan(x−1)⋅dx
After using x=1-ux=1−u an dx=-dudx=−du transforms,
II=int_1^0 arctan(1-u)*(-du)∫01arctan(1−u)⋅(−du)+int_1^0 arctan(-u)*(-du)∫01arctan(−u)⋅(−du)
=int_1^0 arctan(u-1)*du∫01arctan(u−1)⋅du+int_1^0 arctanu*du∫01arctanu⋅du
=-int_0^1 arctanu*du−∫10arctanu⋅du-int_0^1 arctan(u-1)*du∫10arctan(u−1)⋅du
=-int_0^1 arctanx*dx−∫10arctanx⋅dx-int_0^1 arctan(x-1)*dx∫10arctan(x−1)⋅dx
After summing 2 integrals,
2I=02I=0, hence I=0I=0.
