By the following methodology I have determined that the integral of arctan(x)/x^2 dx with lower limit = 1 and upper limit = infinity is divergent. Can somebody verify this? If it is not divergent can you show me that the limit of the function as x-> infinity exists? By using substitution by parts and then partial fractions I found one of the terms to be infinity and concluded that it was, therefore, divergent. I don't have a math scripting program so it is difficult for me to show my work, but here goes:

solve by parts: with u = arctan(x) du = 1/(1+x^2)dx dv= 1/x^2 v= -1/x then limit as x -> inf [-arctan(x)/x + int[1/x(1+x^2)]. Then I use partial fractions on [1/x(1+x^2)] to get int(1/x) + int(-x/(x^2+1)). Integral of 1/x is ln(x). The limit of ln(x) as x-> infinity is infinity. I need go no further, then, because this term makes the limit of the integral indeterminate. So the function is divergent. Is that correct?