What is #int "arccot"x dx#?

Perform an integration by parts

#intuv'=uv-intu'v#

Here,

#u=arc cotx#, #=>#, #u'=-1/(x^2+1)#

#v'=1#, #=>#, #v=x#

The integral is

#I=x arc cotx+int(xdx)/(x^2+1)#

#=x arc cotx+1/2int(2xdx)/(x^2+1)#

#=x arc cotx+1/2ln(x^2+1)+C#

We have,

#int cot^-1x dx#

Integrate by Parts. [#intuvdx = uintvdx - int(d/dx(u)intvdx)#]

#= cot^-1x int dx - int (d/dx(cot^-1x)intdx)dx#

#= xcot^-1x - int(-1/(x^2 + 1) * x)dx#

#= xcot^-1x - intx/(-x^2 - 1)dx#.........................(i)

Now, Let's Solve #int-x/(x^2 + 1) dx#.

#int x/(-x^2 - 1) dx#

Substitute #u = -x^2 - 1#.

That Means, #du = -2x dx rArr dx = -1/(2x) du#

So,

#int x/(-x^2 - 1) dx#

#= int x/u dx#

#= int cancelx/u * (-1/(2cancelx))du#

#= -1/2 int 1/u du#

#= -1/2 ln|u| + C#

#= -1/2 ln|-x^2 - 1| + C#....................(ii)

Now, From (i),

#int cot^-1x dx#

#= xcot^-1x + 1/2ln|-x^2 - 1| + C#

#= xcot^-1x + 1/2ln(x^2 + 1) + C#

And, That's settled.

Hope this helps.

We use the integral of inverse functions theorem:

#intf^-1(x)"d"x=xf^-1(x)+F(f^-1(x))+"c"#

where #F(x)=intf(x)"d"x#

If #f^-1(x)="arccot"x=cot^-1(x)# then #f(x)=cotx# and #F(x)=lnabssinx# (Try to prove this)

So,

#intcot^-1(x)"d"x=xcot^-1(x)-lnsin(cot^-1(x))+"c"#

Note that this is equivalent to the other answer because

#sin(cot^-1(x))=1/csc(cot^-1(x))=1/sqrt(cot^2(cot^-1(x))+1)=1/sqrt(x^2+1)#

so

#-lnsin(cot^-1(x))=-ln(1/sqrt(x^2+1))=-ln((x^2+1)^(-1/2) )=1/2ln(x^2+1)#