Question 1020475
Note first that {{{theta}}} is an angle with its terminal side on the 1st quadrant, and so all its trig functions are positive.

Using the identity {{{sec^2(theta) = 1+tan^2(theta)}}}, we get

{{{tan^2(theta) = (x^2 - 4)/4}}}, or {{{tan(theta) = sqrt(x^2 - 4)/2}}}, where we chose the positive square root because {{{theta}}} is in quadrant 1.

==> {{{ln(abs(sec(theta)+tan(theta))) = ln(abs(x/2+sqrt(x^2 - 4)/2)) = ln(abs(x+sqrt(x^2 - 4))/2) = ln((x+sqrt(x^2 - 4))/2)}}}.
The absolute value bars are not necessary because tangent and secant are positive in quadrant 1.