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.
