Question

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?

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

Answer 1

Given Improper Integral converges to `pi/8`.

Explanation:

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`.