Calculus

put x=2t so u will get half of integrand from 0 to infinity [(t^2+1)/(t^4+1)]* ln(2t) dt . now u cn split ln(2t)dt as ln2+lnt . the first term you cn evaluate you will get{ pie/(2 root of 2)} ln2 and second term u cn solve u will get 0.

$\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \frac { { t }^{ 2 }+1 }{ { t }^{ 4 }+1 } } \ln { (2t)\quad dt\quad =\quad \frac { 1 }{ 2 } } \int _{ 0 }^{ \infty }{ \frac { { t }^{ 2 }+1 }{ { t }^{ 4 }+1 } } \ln { 2dt+\frac { 1 }{ 2 } } \int _{ 0 }^{ \infty }{ \frac { { t }^{ 2 }+1 }{ { t }^{ 4 }+1 } } lnt\quad dt\quad \\ \\ =\frac { \Pi }{ 2\sqrt { 2 } } \ln { 2 } +0.$

sorry for not writing the solution properly. I still dont know how to format properly.