The antiderivative or integral of this is:
#int "arccot"xdx = color(blue)(x"arccot"x + 1/2ln(1+x^2) + C)#
You tend to have to do inverse trig integrals using Integration by Parts (try it on #int arctanxdx# sometime).
Let:
#u = "arccot"x#
#du = -1/(1+x^2)dx#
#dv = 1dx#
#v = x#
#uv - intvdu#
#= x"arccot"x - int (-x)/(1+x^2)dx#
#= x"arccot"x + int x/(1+x^2)dx#
Now, you can determine the integral here using u-substitution. Let:
#u = 1+x^2#
#du = 2xdx#
#=> 1/2 int (2x)/(1+x^2)dx#
#= 1/2 int 1/udu#
#= 1/2ln|u| = 1/2ln|1+x^2| = 1/2ln(1+x^2)#
(since #1+x^2 > 0#, always, for #x in RR#)
Finally, you get your answer as:
#int "arccot"xdx = color(blue)(x"arccot"x + 1/2ln(1+x^2) + C)#
You can remember the derivative of #"arccot"x# by simply remembering the derivative of #arctanx# and multiplying that by #-1#. Simple pattern: take all the non-#co"-"# inverse trig functions and differentiate them. Then multiply by #-1# to get their #co"-"# counterparts, i.e. #"arccsc"#, #"arccot"#, and #"arccos"# vs. #"arcsec"#, #"arctan"#, and #"arcsin"#, respectively.
