Solve the indefinite integral by first substituting #x=t^2#, #dx =2tdt#:
#int (2tlntdt)/(1+t^2)^2 = int (lnsqrtxdx)/(1+x)^2#
Now using the properties of logarithms: #ln sqrtx = 1/2 lnx#
# int (lnsqrtxdx)/(1+x)^2 = 1/2 int (lnx dx)/(1+x)^2#
which we can integrate by parts:
#1/2 int(lnx dx)/(1+x)^2 = 1/2 int lnx d(-1/(1+x))#
#1/2 int(lnx dx)/(1+x)^2 = -1/2 lnx/(1+x) + int dx/(x(1+x))#
and using partial fraction decomposition:
#1/(x(1+x)) = A/x +B/(x+1)#
#1 = Ax+ A +Bx#
#A=1, B=-1#
#1/2 int_1^oo (lnx dx)/(1+x)^2 = -1/2 lnx/(1+x) +1/2 int_1^oo dx/x- 1/2 int_1^oo dx/(1+x) #
#1/2 int_1^oo (lnx dx)/(1+x)^2 = -1/2 lnx/(1+x) +1/2 lnx - 1/2 ln (1+x) +C#
#1/2 int_1^oo (lnx dx)/(1+x)^2 = -1/2 lnx/(1+x) +1/2 ln(x / (1+x) )+C#
and undoing the substitution:
#int (2tlntdt)/(1+t^2)^2 = -1/2 ln t^2/(1+t^2) + 1/2 ln (t^2/(1+t^2))+C#
#int (2tlntdt)/(1+t^2)^2 = ln (t/sqrt(1+t^2)) -ln t/(1+t^2) +C#
in conclusion:
#int_1^oo (2tlntdt)/(1+t^2)^2 = -ln(1/sqrt2) = ln2/2#
