How do you determine if the improper integral converges or diverges #int [ (x arctan x) / (1+x^2)^2 ] dx# from negative 0 to infinity?

Question

Since "negative #0#" is #0#, we deal with the problem as #int_0^oo(xarctanx)/(1+x^2)^2dx#

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Answer 1 · 2016-08-15T18:26:29

Given Improper Integral converges to #pi/8#.

Let #I=int_0^oo (xarctanx)/(1+x^2)^2dx#.

We subst. #x=tant rArr arctanx=t, and, dx=sec^2tdt#.

Also, #(xarctanx)/(1+x^2)^2=((t)(tant))/(1+tan^2t)^2=(t*tant)/sec^4t#

Further, #x=0rArrtant=0rArrt=0, &, xrarroo, trarrpi/2#

#:. I=int_0^(pi/2) (t*tant)/sec^4t*sec^2tdt=int_0^(pi/2) t*sint/cost*cos^2tdt#

#=int_0^(pi/2) tsintcostdt=1/2inttsin(2t)dt#

#=1/2[t(-cos(2t)/2]_0^(pi/2)-1/2int_0^(pi/2)1*(-cos(2t)/2)dt#

#=-1/4[pi/2*cos(2*pi/2)-0]+1/4int_0^(pi/2)cos(2t)dt#

#=-1/4(-pi/2)+1/4[sin(2t)/2]_0^(pi/2)#

#=pi/8+1/8[sinpi-sin0]#

#=pi/8#.

Hence, the given Improper Integral #I# converges to #pi/8#.