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