#int arctan(1/x) dx#
Let #theta = arctan(1/x)#.
This makes #tan theta = 1/x#, so #cot theta = x#.
Furthermore, #dx = -csc^2 theta " " d theta#
The integral becomes:
#int theta (-csc^2 theta) d theta#
Let #u = theta# and #dv = (-csc^2 theta) d theta#
So #du = d theta# and #v = cot theta#
#uv-int v du = theta cot theta - int cot theta d theta#
The integral can be found by substitution. We get
#theta cot theta -ln abs sin theta +C#
Using #cot theta = x# and some trigonometry, we sind #sin theta = 1/sqrt(x^2+1)#
Therefore
#int arctan(1/x) dx = x arctan (1/x)-ln(1/sqrt(x^2+1))+C#
# = x arctan(1/x)+1/2 ln(x^2+1) +C#
