Evaluate the integral of arctan(1/x).

Let u=arctan (1/x),
du = 1/(1+(1/x)²) * (-1/x²) dx = -1/(x²+1) dx , dv = dx, v = x.

Then,
[1, √3]∫arctan (1/x) dx 
x arctan (1/x) |[1, √3] + [1, √3]∫x/(x²+1) dx 
√3 arctan (1/√3) - 1 arctan 1 + [1, √3]∫x/(x²+1) dx 
π√3/6 - π/4 + [1, √3]∫x/(x²+1) dx