Mark M. answered 02/29/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Let u = Arctan(1/x) and dv = dx.
Then du = [1 / (1 + (1/x2))] (-1/x2)dx and v = x.
Simplifying, we get du = -1 / (1 + x2) dx and v = x.
So, ∫Arctan(1/x)dx = xArctan(1/x) + ∫[x / (1 + x2) ]dx = xArctan(1/x) + (1/2) ln(1+x2) + C
Evaluating from 1 to √3, we have: √3Arctan(1/√3) + (1/2)ln4 - Arctan(1) - (1/2)ln2
= √3(π/6) - π/4 + (1/2) ln2