For x>0, let f(x)=∫x1log t1+t dt. Then f(x)+f(1x) is equal to
12(logx)2
f(1x)=∫1x1log t1+t dt
Put t=1u, so that dt=(−1u2) du, then
f(1x)=∫x1log(1u)1+1u(−1u2) du
=∫x1log uu(1+u) du=∫x1log tt(1+t) dt
∴f(x)+f(1x)=∫x1(1+1t)log t1+t dt
=∫x1log tt dt
=12 (log t)2]x1=12(log x)2